Rd Sharma Solutions for Class 10 Math Chapter 14 Co Ordinate Geometry are provided here with simple step-by-step explanations. These solutions for Co Ordinate Geometry are extremely popular among Class 10 students for Math Co Ordinate Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Book of Class 10 Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Solutions. All Rd Sharma Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### Page No 14.14:

#### Question 1:

Find the distance between the following pair of points:

(a) (−6, 7) and (−1, −5)

(b) (*a*+*b*, *b*+*c*) and (*a*−*b*, *c*−*b*)

(c) (*a*sinα, −*b*cosα) and (−*a*cos α, *b*sin α)

(d) (*a*, 0) and (0, *b*)

#### Answer:

The distance *d* between two points and is given by the formula

(i) The two given points are (−6, 7) and (−1, −5)

The distance between these two points is

Hence the distance is.

(ii) The two given points are

The distance between these two points is

Hence the distance is.

(iii) The two given points are and

The distance between these two points is

Hence the distance is.

(iv) The two given points are (*a, *0) and (0*, b*)

The distance between these two points is

Hence the distance is.

#### Page No 14.14:

#### Question 2:

Find the value of *a* when the distance between the points (3, *a*) and (4, 1) is $\sqrt{10}$.

#### Answer:

The distance *d* between two points and is given by the formula

The distance between two points (3*, a*) and (4*, *1) is given as. Substituting these values in the formula for distance between two points we have,

Now, squaring the above equation on both sides of the equals sign

Thus we arrive at a quadratic equation. Let us solve this now,

The roots of the above quadratic equation are thus 4 and −2.

Thus the value of ‘*a*’ could either be.

#### Page No 14.14:

#### Question 3:

If the points (2, 1) and (1, −2) are equidistant from the point (*x*, *y*), show that *x* + 3*y* = 0.

#### Answer:

The distance *d* between two points and is given by the formula

Here it is said that the points (2*, *1) and (1*, **−*2) are equidistant from (*x, y*).

Let be the distance between (2*, *1)* *and (*x*,* y*).

Let be the distance between (1, −2)* *and (*x, y*).

So, using the distance formula for both these pairs of points we have

Now since both these distances are given to be the same, let us equate both and

Squaring on both sides we have,

Hence we have proved that when the points (2*, *1) and (1*,−*2) are equidistant from (*x, y*) we have .

#### Page No 14.14:

#### Question 4:

Find the values of *x*,* y* if the distances of the point (*x*, *y*) from (−3, 0) as well as from (3, 0) are 4.

#### Answer:

The distance *d* between two points and is given by the formula

It is said that (*x, y*) is equidistant from both (−3,0) and (3,0).

Let be the distance between (*x, y*) and (−3,0).

Let be the distance between (*x, y*) and (3,0).

So, using the distance formula for both these pairs of points we have

Now since both these distances are given to be the same, let us equate both and .

Squaring on both sides we have,

It is also said that the value of both and is 4 units.

Substituting the value of ‘*x*’ in the equation for either or we can get the value of ‘*y*’.

Squaring on both sides of the equation we have,

Hence the values of ‘*x*’ and ‘*y*’ are.

#### Page No 14.14:

#### Question 5:

The length of a line segment is of 10 units and the coordinates of one end-point are (2, −3). If the abscissa of the other end is 10, find the ordinate of the other end.

#### Answer:

The distance *d* between two points and is given by the formula

Here it is given that one end of a line segment has co−ordinates (2*,−*3). The abscissa of the other end of the line segment is given to be 10. Let the ordinate of this point be ‘*y*’.

So, the co−ordinates of the other end of the line segment is (10*, y*).

The distance between these two points is given to be 10 units.

Substituting these values in the formula for distance between two points we have,

Squaring on both sides of the equation we have,

We have a quadratic equation for ‘*y*’. Solving for the roots of this equation we have,

The roots of the above equation are ‘*−*9’ and ‘3’

Thus the ordinates of the other end of the line segment could be.

#### Page No 14.14:

#### Question 6:

Show that the points (−4, −1), (−2, −4) (4, 0) and (2, 3) are the vertices points of a rectangle.

#### Answer:

The distance *d* between two points and is given by the formula

In a rectangle, the opposite sides are equal in length. The diagonals of a rectangle are also equal in length.

Here the four points are *A*(*−*4*,**−*1)*, B*(*−*2*,**−*4), *C*(4*,*0) and *D*(2*,*3).

First let us check the length of the opposite sides of the quadrilateral that is formed by these points.

We have one pair of opposite sides equal.

Now, let us check the other pair of opposite sides.

The other pair of opposite sides are also equal. So, the quadrilateral formed by these four points is definitely a parallelogram.

For a parallelogram to be a rectangle we need to check if the diagonals are also equal in length.

Now since the diagonals are also equal we can say that the parallelogram is definitely a rectangle.

Hence we have proved that the quadrilateral formed by the four given points is a.

#### Page No 14.14:

#### Question 7:

Show that the points *A* (1, −2), *B* (3, 6), *C* (5, 10) and *D* (3, 2) are the vertices of a parallelogram.

#### Answer:

The distance *d* between two points and is given by the formula

In a parallelogram the opposite sides are equal in length.

Here the four points are *A*(1*,** −*2)*, B*(3*,* 6), *C*(5*,* 10) and *D*(3*, *2).

Let us check the length of the opposite sides of the quadrilateral that is formed by these points.

We have one pair of opposite sides equal.

Now, let us check the other pair of opposite sides.

The other pair of opposite sides is also equal. So, the quadrilateral formed by these four points is definitely a parallelogram.

Hence we have proved that the quadrilateral formed by the given four points is a .

#### Page No 14.14:

#### Question 8:

Prove that the points *A*(1, 7), *B* (4, 2), *C*(−1, −1) *D* (−4, 4) are the vertices of a square.

#### Answer:

The distance *d* between two points and is given by the formula

In a square all the sides are equal in length. Also, the diagonals are equal in length in a square.

Here the four points are *A*(1*, *7)*, B*(4*,* 2), *C*(*−*1*,** −*1) and *D*(*−*4*,* 4).

First let us check if all the four sides are equal.

Since all the sides of the quadrilateral are the same it is a rhombus.

For the rhombus to be a square the diagonals also have to be equal to each other.

Since the diagonals of the rhombus are also equal to each other the rhombus is a square.

Hence the quadrilateral formed by the given points is a.

#### Page No 14.14:

#### Question 9:

Prove that the points (3, 0), (6, 4) and (−1, 3) are vertices of a right-angled isosceles triangle.

#### Answer:

The distance *d* between two points and is given by the formula

In an isosceles triangle there are two sides which are equal in length.

Here the three points are *A*(3*,* 0)*, B*(6*, *4) and *C*(*−*1*, *3).

Let us check the length of the three sides of the triangle.

Here, we see that two sides of the triangle are equal. So the triangle formed should be an isosceles triangle.

We can also observe that

Hence proved that the triangle formed by the three given points is an.

#### Page No 14.15:

#### Question 10:

Prove that (2, −2) (−2, 1) and (5, 2) are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse.

#### Answer:

The distance *d* between two points and is given by the formula

In a right-angled triangle, by Pythagoras theorem, the square of the longest side is equal to the sum of squares of the other two sides in the triangle.

Here the three points are *A*(2*,**−*2)*, B*(*−*2*,*1) and *C*(5*,*2).

Let us find out the lengths of all the sides of the triangle.

Here we have,

Since the square of the longest side is equal to the sum of squares of the other two sides the given triangle is a .

In a right angled triangle the area of the triangle ‘*A*’ is given by,

In a right angled triangle the sides containing the right angle will not be the longest side.

Hence the area of the given right angled triangle is,

Hence the area of the given right-angled triangle is.

In a right-angled triangle the hypotenuse will be the longest side. Here the longest side is ‘*BC*’.

Hence the hypotenuse of the given right-angled triangle is

#### Page No 14.15:

#### Question 11:

Prove that the points (2*a*, 4*a*), (2*a*, 6*a*) and (2*a*+$\sqrt{3}a$, 5*a*) are the vertices of an equilateral triangle.

#### Answer:

The distance *d* between two points and is given by the formula

In an equilateral triangle all the sides have equal length.

Here the three points are, and.

Let us now find out the lengths of all the three sides of the given triangle.

Since all the three sides have equal lengths the triangle has to be an equilateral triangle.

#### Page No 14.15:

#### Question 12:

Prove that the points (2,3), (−4, −6) and (1, 3/2) do not form a triangle.

#### Answer:

The distance *d* between two points and is given by the formula

In any triangle the sum of lengths of any two sides need to be greater than the third side.

Here the three points are, and

Let us now find out the lengths of all the three sides of the given triangle.

Here we see that,

This is in violation of the basic property of any triangle to exist. Therefore these points cannot give rise to a triangle.

Hence we have proved that the given three points do not form a triangle.

#### Page No 14.15:

#### Question 13:

An equilateral triangle has two vertices at the points (3, 4) and (−2, 3), find the coordinates of the third vertex.

#### Answer:

The distance *d* between two points and is given by the formula

In an equilateral triangle all the sides are of equal length.

Here we are given that *A *(3*, *4) and *B *(*−*2*, *3) are two vertices of an equilateral triangle. Let *C*(*x, y*) be the third vertex of the equilateral triangle.

First let us find out the length of the side of the equilateral triangle.

Hence the side of the equilateral triangle measures units.

Now, since it is an equilateral triangle, all the sides need to measure the same length.

Hence we have

Equating both these equations we have,

Squaring on both sides we have,

From the above equation we have,

Substituting this and the value of the side of the triangle in the equation for one of the sides we have,

Squaring on both sides,

Now we have a quadratic equation for ‘*x*’. Solving for the roots of this equation,

We know that. Substituting the value of ‘*x*’ we have,

Hence the two possible values of the third vertex are.

#### Page No 14.15:

#### Question 14:

Show that the quadrilateral whose vertices are (2, −1), (3, 4) (−2, 3) and (−3,−2) is a rhombus.

#### Answer:

The distance *d* between two points and is given by the formula

In a rhombus all the sides are equal in length.

Here the four points are *A *(2*, **−*1)*, B *(3*, *4), *C *(*−*2*, *3) and *D *(*−*3*, **−*2).

First let us check if all the four sides are equal.

Here, we see that all the sides are equal, so it has to be a rhombus.

#### Page No 14.15:

#### Question 15:

Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3.

#### Answer:

The distance *d* between two points and is given by the formula

In an isosceles triangle two sides will be of equal length.

Here two vertices of the triangle is given as *A *(2*, *0) and *B *(2*, *5). Let the third side of the triangle be *C*(*x, y*)

It is given that the length of the equal sides is 3 units.

Let us now find the length of the side in which both the vertices are known.

So, now we know that the side ‘*AB*’ is not one of the equal sides of the isosceles triangle.

So, we have

Equating these two equations we have,

Squaring on both sides of the equation we have,

We know that the length of the equal sides is 3 units. So substituting the value of ‘*y*’ in equation for either ‘*AC*’ or ‘*BC*’ we can get the value of ‘*x*’.

Squaring on both sides,

We have a quadratic equation for ‘*x*’. Solving for roots of the above equation we have,

Hence the possible co−ordinates of the third vertex of the isosceles triangle are.

#### Page No 14.15:

#### Question 16:

Which point on x-axis is equidistant from (5, 9) and (−4, 6) ?

#### Answer:

The distance *d* between two points and is given by the formula

Here we are to find out a point on the *x*−axis which is equidistant from both the points *A *(5*, *9) and *B *(*−*4*, *6)

Let this point be denoted as *C*(*x, y*)

Since the point lies on the *x*-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘*A*’ and ‘*B*’ to ‘*C*’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the *x*-axis which lies at equal distances from the mentioned points is.

#### Page No 14.15:

#### Question 17:

Prove that the points (−2, 5), (0, 1) and (2, −3) are collinear.

#### Answer:

The distance *d* between two points and is given by the formula

For three points to be collinear the sum of distances between two pairs of points should be equal to the third pair of points.

The given points are *A *(*−*2*, *5), *B *(0*, *1) and *C *(2*, **−*3)

Let us find the distances between the possible pairs of points.

We see that

Since sum of distances between two pairs of points equals the distance between the third pair of points the three points must be collinear.

Hence we have proved that the three given points are.

#### Page No 14.15:

#### Question 18:

The coordinates of the point *P* are (−3, 2). Find the coordinates of the point *Q* which lies on the line joining *P* and origin such that OP = OQ.

#### Answer:

If and are given as two points, then the co-ordinates of the midpoint of the line joining these two points is given as

It is given that the point ‘*P*’ has co-ordinates (*−*3*, *2)

Here we are asked to find out the co-ordinates of point ‘*Q*’ which lies along the line joining the origin and point ‘*P*’. Thus we can see that the points ‘*P*’, ‘*Q*’ and the origin are collinear.

Let the point ‘*Q*’ be represented by the point (*x, y*)

Further it is given that the

This implies that the origin is the midpoint of the line joining the points ‘*P*’ and ‘*Q*’.

So we have that

Substituting the values in the earlier mentioned formula we get,

Equating individually we have, and.

Thus the co−ordinates of the point ‘*Q*’ is

#### Page No 14.15:

#### Question 19:

Which point on *y*-axis is equidistant from (2, 3) and (−4, 1)?

#### Answer:

The distance *d* between two points and is given by the formula

Here we are to find out a point on the *y*-axis which is equidistant from both the points *A *(2*, *3) and *B *(*−*4*, *1).

Let this point be denoted as *C*(*x, y*).

Since the point lies on the *y*-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘*A*’ and ‘*B*’ to ‘*C*’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the *y*-axis which lies at equal distances from the mentioned points is.

#### Page No 14.15:

#### Question 20:

The three vertices of a parallelogram are (3, 4) (3, 8) and (9, 8). Find the fourth vertex.

#### Answer:

We are given three vertices of a parallelogram A(3, 4), B(3, 8) and C(9, 8).

We know that diagonals of a parallelogram bisect each other. Let the fourth vertex be D(*x, y*).

Mid point of BD = $\left(\frac{x+3}{2},\frac{y+8}{2}\right)$

Mid point of AC = $\left(\frac{9+3}{2},\frac{4+8}{2}\right)$ = $\left(6,6\right)$

Since the mid point of BD = mid point of AC

So, $\left(\frac{x+3}{2},\frac{y+8}{2}\right)$ = (6, 6)

Thus, *x* = 9, *y* = 4.

So, the fourth vertex is (9, 4).

#### Page No 14.15:

#### Question 21:

Find the circumcentre of the triangle whose vertices are (−2, −3), (−1, 0), (7, −6).

#### Answer:

The distance *d* between two points and is given by the formula

The circumcentre of a triangle is the point which is equidistant from each of the three vertices of the triangle.

Here the three vertices of the triangle are given to be *A*(*−*2*.−*3), *B*(*−*1*,*0) and *C*(7*,**−*6).

Let the circumcentre of the triangle be represented by the point *R*(*x, y*).

So we have

Equating the first pair of these equations we have,

Squaring on both sides of the equation we have,

Equating another pair of the equations we have,

Squaring on both sides of the equation we have,

Now we have two equations for ‘*x*’ and ‘*y*’, which are

From the second equation we have. Substituting this value of ‘*y*’ in the first equation we have,

Therefore the value of ‘*y*’ is,

Hence the co−ordinates of the circumcentre of the triangle with the given vertices are.

#### Page No 14.15:

#### Question 22:

Find the angle subtended at the origin by the line segment whose end points are (0, 100) and (10, 0).

#### Answer:

The distance *d* between two points and is given by the formula

In a right angled triangle the angle opposite the hypotenuse subtends an angle of.

Here let the given points be *A*(0*,*100)*, B*(10*,*0). Let the origin be denoted by *O*(0*,*0).

Let us find the distance between all the pairs of points

Here we can see that.

So, is a right angled triangle with ‘*AB*’ being the hypotenuse. So the angle opposite it has to be. This angle is nothing but the angle subtended by the line segment ‘*AB*’ at the origin.

Hence the angle subtended at the origin by the given line segment is.

#### Page No 14.15:

#### Question 23:

Find the centre of the circle passing through (5, −8), (2, −9) and (2, 1).

#### Answer:

The distance *d* between two points and is given by the formula

The centre of a circle is at equal distance from all the points on its circumference.

Here it is given that the circle passes through the points *A*(5*,**−*8), *B*(2*,**−*9) and *C*(2*,*1).

Let the centre of the circle be represented by the point *O*(*x, y*).

So we have

Equating the first pair of these equations we have,

Squaring on both sides of the equation we have,

Equating another pair of the equations we have,

Squaring on both sides of the equation we have,

Now we have two equations for ‘*x*’ and ‘*y*’, which are

From the second equation we have. Substituting this value of ‘*y*’ in the first equation we have,

Therefore the value of ‘*y*’ is,

Hence the co-ordinates of the centre of the circle are.

#### Page No 14.15:

#### Question 24:

Find the value of *k*, if the point *P* (0, 2) is equidistant from (3, *k*) and (*k*, 5).

#### Answer:

The distance *d* between two points and is given by the formula

It is said that *P*(0*,*2) is equidistant from both *A*(3*,k*) and *B*(*k,*5).

So, using the distance formula for both these pairs of points we have

Now since both these distances are given to be the same, let us equate both.

Squaring on both sides we have,

Hence the value of ‘*k*’ for which the point ‘*P*’ is equidistant from the other two given points is.

#### Page No 14.15:

#### Question 25:

If two opposite vertices of a square are (5, 4) and (1, −6), find the coordinates of its remaining two vertices.

#### Answer:

The distance *d* between two points and is given by the formula

In a square all the sides are of equal length. The diagonals are also equal to each other. Also in a square the diagonal is equal to times the side of the square.

Here let the two points which are said to be the opposite vertices of a diagonal of a square be *A*(5*,*4)* *and *C*(1*,**−*6).

Let us find the distance between them which is the length of the diagonal of the square.

Now we know that in a square,

Substituting the value of the diagonal we found out earlier in this equation we have,

Now, a vertex of a square has to be at equal distances from each of its adjacent vertices.

Let *P*(*x, y*) represent another vertex of the same square adjacent to both ‘*A*’ and ‘*C*’.

But these two are nothing but the sides of the square and need to be equal to each other.

Squaring on both sides we have,

From this we have,

Substituting this value of ‘*x*’ and the length of the side in the equation for ‘*AP*’ we have,

Squaring on both sides,

We have a quadratic equation. Solving for the roots of the equation we have,

The roots of this equation are −3 and 1.

Now we can find the respective values of ‘*x*’ by substituting the two values of ‘*y*’

When

When

Therefore the other two vertices of the square are.

#### Page No 14.15:

#### Question 26:

Show that the points (−3, 2), (−5,−5), (2, −3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus.

#### Answer:

The distance *d* between two points and is given by the formula

In a rhombus all the sides are equal in length. And the area ‘*A*’ of a rhombus is given as

Here the four points are *A*(*−*3*,*2)*, B*(*−*5*,**−*5), *C*(2*,**−*3) and *D*(4*,*4)

First let us check if all the four sides are equal.

Here, we see that all the sides are equal, so it has to be a rhombus.

Hence we have proved that the quadrilateral formed by the given four vertices is a.

Now let us find out the lengths of the diagonals of the rhombus.

Now using these values in the formula for the area of a rhombus we have,

Thus the area of the given rhombus is.

#### Page No 14.15:

#### Question 27:

Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (−1, −6) and (4, −1). Also, find its circumradius.

#### Answer:

The distance *d* between two points and is given by the formula

The circumcentre of a triangle is the point which is equidistant from each of the three vertices of the triangle.

Here the three vertices of the triangle are given to be *A*(3*,*0), *B*(*−*1*,**−*6) and *C*(4*,**−*1)

Let the circumcentre of the triangle be represented by the point *R*(*x, y*).

So we have

Equating the first pair of these equations we have,

Squaring on both sides of the equation we have,

Equating another pair of the equations we have,

Squaring on both sides of the equation we have,

Now we have two equations for ‘*x*’ and ‘*y*’, which are

From the second equation we have. Substituting this value of ‘*y*’ in the first equation we have,

Therefore the value of ‘*y*’ is,

Hence the co-ordinates of the circumcentre of the triangle with the given vertices are.

The length of the circumradius can be found out substituting the values of ‘*x*’ and ‘*y*’ in ‘*AR*’

Thus the circumradius of the given triangle is.

#### Page No 14.15:

#### Question 28:

Find a point on the *x*-axis which is equidistant from the points (7, 6) and (−3, 4).

#### Answer:

The distance *d* between two points and is given by the formula

Here we are to find out a point on the *x*−axis which is equidistant from both the points *A*(7*,*6) and *B*(*−*3*,*4).

Let this point be denoted as *C*(*x, y*).

Since the point lies on the *x*-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘*A*’ and ‘*B*’ to ‘*C*’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the *x*-axis which lies at equal distances from the mentioned points is.

#### Page No 14.15:

#### Question 29:

(i) Show that the points *A*(5, 6), *B*(1, 5), *C*(2, 1) and *D*(6,2) are the vertices of a square.

(ii) Prove hat the points *A* (2, 3) *B*(−2,2) *C*(−1,−2), and *D*(3, −1) are the vertices of a square *ABCD*.

#### Answer:

The distance *d* between two points and is given by the formula

In a square all the sides are equal to each other. And also the diagonals are also equal to each other.

Here the four points are *A*(5*,*6)*, B*(1*,*5), *C*(2*,*1) and *D*(6*,*2).

First let us check if all the four sides are equal.

Here, we see that all the sides are equal, so it has to be a rhombus.

Now let us find out the lengths of the diagonals of this rhombus.

Now since the diagonals of the rhombus are also equal to each other this rhombus has to be a square.

Hence we have proved that the quadrilateral formed by the given four points is a.

#### Page No 14.15:

#### Question 30:

Find the point on x-axis which is equidistant from the points (−2, 5) and (2,−3).

#### Answer:

The distance *d* between two points and is given by the formula

Here we are to find out a point on the *x-*axis which is equidistant from both the points *A*(*−*2*,*5) and *B*(2*,**−*3)

Let this point be denoted as *C*(*x, y*).

Since the point lies on the *x*-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘*A*’ and ‘*B*’ to ‘*C*’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the *x*-axis which lies at equal distances from the mentioned points is.

#### Page No 14.15:

#### Question 31:

Find the value of *x* such that *PQ* = *QR* where the coordinates of *P*,* Q* and *R* are (6, −1) , (1, 3) and (*x*, 8) respectively.

#### Answer:

The distance *d* between two points and is given by the formula

The three given points are *P*(6*,**−*1)*, Q*(1*,*3)* *and *R*(*x,*8).

Now let us find the distance between ‘*P*’ and ‘*Q*’.

Now, let us find the distance between ‘*Q*’ and ‘*R*’.

It is given that both these distances are equal. So, let us equate both the above equations,

Squaring on both sides of the equation we get,

Now we have a quadratic equation. Solving for the roots of the equation we have,

Thus the roots of the above equation are 5 and −3.

Hence the values of ‘*x*’ are.

#### Page No 14.15:

#### Question 32:

Prove that the points (0, 0), (5, 5) and (−5, 5) are the vertices of a right isosceles triangle.

#### Answer:

The distance *d* between two points and is given by the formula

In an isosceles triangle there are two sides which are equal in length.

By Pythagoras Theorem in a right-angled triangle the square of the longest side will be equal to the sum of squares of the other two sides.

Here the three points are *A*(0*,*0)*, B*(5*,*5) and *C*(*−*5*,*5).

Let us check the length of the three sides of the triangle.

Here, we see that two sides of the triangle are equal. So the triangle formed should be an isosceles triangle.

Further it is seen that

If in a triangle the square of the longest side is equal to the sum of squares of the other two sides then the triangle has to be a right-angled triangle.

Hence proved that the triangle formed by the three given points is an.

#### Page No 14.15:

#### Question 33:

If the point *P*(*x*, *y*) is equidistant from the points *A*(5, 1) and *B* (1, 5), prove that *x* = *y*.

#### Answer:

The distance *d* between two points and is given by the formula

The three given points are* P*(*x, y*)*, A*(5*,*1)* *and *B*(1*,*5).

Now let us find the distance between ‘*P*’ and ‘*A*’.

Now, let us find the distance between ‘*P*’ and ‘*B*’.

$PB=\sqrt{{\left(x-1\right)}^{2}+{\left(y-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}$

It is given that both these distances are equal. So, let us equate both the above equations,

PA = PB

$\sqrt{{\left(x-5\right)}^{2}+{\left(y-1\right)}^{2}}=\sqrt{{\left(x-1\right)}^{2}+{\left(y-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}$

Squaring on both sides of the equation we get,

${\left(x-5\right)}^{2}+{\left(y-1\right)}^{2}={\left(x-1\right)}^{2}+{\left(y-5\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+25-10x+{y}^{2}+1-2y={x}^{2}+1-2x+{y}^{2}+25-10y\phantom{\rule{0ex}{0ex}}\Rightarrow 26-10x-2y=26-10y-2x\phantom{\rule{0ex}{0ex}}\Rightarrow 10y-2y=10x-2x\phantom{\rule{0ex}{0ex}}\Rightarrow 8y=8x\phantom{\rule{0ex}{0ex}}\Rightarrow y=x\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence we have proved that *x** *= *y*.

#### Page No 14.15:

#### Question 34:

If *Q* (0,1) is equidistant from *P* (5, −3) and *R* (*x*, 6), find the values of *x*. Also, find the distances *QR* and *PR*.

#### Answer:

The distance *d* between two points and is given by the formula

The three given points are* Q*(0*,*1)*, P*(5*,**−*3)* *and *R*(*x,*6).

Now let us find the distance between ‘*P*’ and ‘*Q*’.

Now, let us find the distance between ‘*Q*’ and ‘*R*’.

It is given that both these distances are equal. So, let us equate both the above equations,

Squaring on both sides of the equation we get,

Hence the values of ‘*x*’ are.

Now, the required individual distances,

Hence the length of ‘*QR*’ is.

For ‘*PR*’ there are two cases. First when the value of ‘*x*’ is 4,

Then when the value of ‘*x*’ is −4,

Hence the length of ‘*PR*’ can beunits

#### Page No 14.15:

#### Question 35:

Find the values of *y* for which the distance between the points *P* (2, −3) and *Q*(10,*y*) is 10 units.

#### Answer:

The distance *d* between two points and is given by the formula

The distance between two points *P*(2*,**−*3) and *Q*(10*,y*) is given as 10 units. Substituting these values in the formula for distance between two points we have,

Now, squaring the above equation on both sides of the equals sign

Thus we arrive at a quadratic equation. Let us solve this now,

The roots of the above quadratic equation are thus 3 and −9.

Thus the value of ‘*y*’ could either be.

#### Page No 14.15:

#### Question 36:

Find the centre of the circle passing through (6, −6), (3, −7) and (3, 3).

#### Answer:

The distance *d* between two points and is given by the formula

The centre of a circle is at equal distance from all the points on its circumference.

Here it is given that the circle passes through the points *A*(6*,**−*6), *B*(3*,**−*7) and *C*(3*,*3).

Let the centre of the circle be represented by the point *O*(*x, y*).

So we have

Equating the first pair of these equations we have,

Squaring on both sides of the equation we have,

Equating another pair of the equations we have,

Squaring on both sides of the equation we have,

Now we have two equations for ‘*x*’ and ‘*y*’, which are

From the second equation we have. Substituting this value of ‘*y*’ in the first equation we have,

Therefore the value of ‘*y*’ is,

Hence the co-ordinates of the centre of the circle are.

#### Page No 14.15:

#### Question 37:

Two opposite vertices of a square are (−1, 2) and (3, 2). Find the coordinates of other two vertices.

#### Answer:

The distance *d* between two points and is given by the formula

In a square all the sides are of equal length. The diagonals are also equal to each other. Also in a square the diagonal is equal to times the side of the square.

Here let the two points which are said to be the opposite vertices of a diagonal of a square be *A*(*−*1*,*2)* *and *C*(3*,*2).

Let us find the distance between them which is the length of the diagonal of the square.

Now we know that in a square,

Substituting the value of the diagonal we found out earlier in this equation we have,

Now, a vertex of a square has to be at equal distances from each of its adjacent vertices.

Let *P*(*x, y*) represent another vertex of the same square adjacent to both ‘*A*’ and ‘*C*’.

But these two are nothing but the sides of the square and need to be equal to each other.

Squaring on both sides we have,

Substituting this value of ‘*x*’ and the length of the side in the equation for ‘*AP*’ we have,

Squaring on both sides,

We have a quadratic equation. Solving for the roots of the equation we have,

The roots of this equation are 0 and 4.

Therefore the other two vertices of the square are.

#### Page No 14.16:

#### Question 44:

If a point A(0, 2) is equidistant from the points B(3, *p*) and C(*p*, 5), then find the value of *p*. [CBSE 2012, 2013]

#### Answer:

It is given that A(0, 2) is equidistant from the points B(3, *p*) and C(*p*, 5).

∴ AB = AC

$\Rightarrow \sqrt{{\left(3-0\right)}^{2}+{\left(p-2\right)}^{2}}=\sqrt{{\left(p-0\right)}^{2}+{\left(5-2\right)}^{2}}$ (Distance formula)

Squaring on both sides, we get

$9+{p}^{2}-4p+4={p}^{2}+9\phantom{\rule{0ex}{0ex}}\Rightarrow -4p+4=0\phantom{\rule{0ex}{0ex}}\Rightarrow p=1$

Thus, the value of *p* is 1.

#### Page No 14.16:

#### Question 45:

Prove that the points (7, 10), (−2, 5) and (3, −4) are the vertices of an isosceles right triangle. [CBSE 2013]

#### Answer:

Let the given points be A(7, 10), B(−2, 5) and C(3, −4).

Using distance formula, we have

$\mathrm{AB}=\sqrt{{\left(-2-7\right)}^{2}+{\left(5-10\right)}^{2}}=\sqrt{{\left(-9\right)}^{2}+{\left(-5\right)}^{2}}=\sqrt{81+25}=\sqrt{106}$ units

$\mathrm{BC}=\sqrt{{\left[3-\left(-2\right)\right]}^{2}+{\left(-4-5\right)}^{2}}=\sqrt{{5}^{2}+{\left(-9\right)}^{2}}=\sqrt{25+81}=\sqrt{106}$ units

$\mathrm{CA}=\sqrt{{\left(3-7\right)}^{2}+{\left(-4-10\right)}^{2}}=\sqrt{{\left(-4\right)}^{2}+{\left(-14\right)}^{2}}=\sqrt{16+196}=\sqrt{212}$ units

Thus, AB = BC = $\sqrt{106}$ units

∴ ∆ABC is an isosceles triangle.

Also,

AB^{2} + BC^{2 }= 106 + 106 = 212

and CA^{2} = 212

∴ AB^{2} + BC^{2 }= CA^{2}

So, ∆ABC is right angled at B. (Converse of Pythagoras theorem)

Hence, the given points are the vertices of an isosceles right triangle.

#### Page No 14.16:

#### Question 38:

Name the quadrilateral formed, if any, by the following points, and given reasons for your answers:

(i) A(−1,−2) B(1, 0), C (−1, 2), D(−3, 0)

(ii) A(−3, 5) B(3, 1), C (0, 3), D(−1, −4)

(iii) A(4, 5) B(7, 6), C (4, 3), D(1, 2)

#### Answer:

(i) A (−1,−2) , B(1,0), C(−1,2), D(−3,0)

Let A, B, C and D be the four vertices of the quadrilateral ABCD.

We know the distance between two points Pand Qis given by distance formula:

Hence

Similarly,

Similarly,

Also,

Hence from above we see that all the sides of the quadrilateral are equal. Hence it is a square.

(ii) A (−3,5) , B(3,1), C(0,3), D(−1,−4)

Let A, B, C and D be the four vertices of the quadrilateral ABCD.

We know the distance between two points Pand Qis given by distance formula:

Hence

Similarly,

Similarly,

Also,

Hence from the above we see that it is not a quadrilateral

(iii) A (4, 5), B (7,6), C(4,3), D(1,2)

Let A, B, C and D be the four vertices of the quadrilateral ABCD.

We know the distance between two points Pand Qis given by distance formula:

Hence

Similarly,

Similarly,

Also,

Hence from above we see that

AB = CD and BC = DA

Here opposite sides of the quadrilateral is equal. Hence it is a parallelogram.

#### Page No 14.16:

#### Question 39:

Find the equation of the perpendicular bisector of the line segment joining points (7, 1) and (3,5).

#### Answer:

TO FIND: The equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5)

Let P(*x*, *y*) be any point on the perpendicular bisector of AB. Then,

PA=PB

Hence the equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5) is

#### Page No 14.16:

#### Question 40:

Prove that the points (3, 0), (4, 5), (−1, 4) and (−2 −1), taken in order, form a rhombus. Also, find its area.

#### Answer:

The distance *d* between two points and is given by the formula

In a rhombus all the sides are equal in length. And the area ‘*A*’ of a rhombus is given as

Here the four points are *A*(3*,*0)*, B*(4*,*5), *C*(*−*1*,*4) and *D*(*−*2*,**−*1).

First let us check if all the four sides are equal.

Here, we see that all the sides are equal, so it has to be a rhombus.

Hence we have proved that the quadrilateral formed by the given four vertices is a.

Now let us find out the lengths of the diagonals of the rhombus.

Now using these values in the formula for the area of a rhombus we have,

Thus the area of the given rhombus is.

#### Page No 14.16:

#### Question 41:

In the seating arrangement of desks in a classroom three students Rohini, Sandhya and Bina are seated at A(3, 1), B(6, 4), and C(8, 6). Do you think they are seated in a line?

#### Answer:

The distance *d* between two points and is given by the formula

For three points to be collinear the sum of distances between any two pairs of points should be equal to the third pair of points.

The given points are *A*(3*,*1), *B*(6*,*4) and *C*(8*,*6).

Let us find the distances between the possible pairs of points.

We see that.

Since sum of distances between two pairs of points equals the distance between the third pair of points the three points must be collinear.

Hence, the three given points are.

#### Page No 14.16:

#### Question 42:

Find a point on *y*-axis which is equidistant form the points (5, −2) and (−3, 2).

#### Answer:

The distance *d* between two points and is given by the formula

Here we are to find out a point on the y−axis which is equidistant from both the points *A*(5*,**−*2) and *B*(*−*3*,*2).

Let this point be denoted as *C*(*x, y*).

Since the point lies on the *y*-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘*A*’ and ‘*B*’ to ‘*C*’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the *y*-axis which lies at equal distances from the mentioned points is.

#### Page No 14.16:

#### Question 43:

Find a relation between *x* and *y* such that the point (*x*, *y*) is equidistant from the points (3, 6) and (−3, 4).

#### Answer:

The distance *d* between two points and is given by the formula

Let the three given points be* P*(*x, y*)*, A*(3*,*6)* *and *B*(*−*3*,*4).

Now let us find the distance between ‘*P*’ and ‘*A*’.

Now, let us find the distance between ‘*P*’ and ‘*B*’.

It is given that both these distances are equal. So, let us equate both the above equations,

Squaring on both sides of the equation we get,

Hence the relationship between ‘*x*’ and ‘*y*’ based on the given condition is.

#### Page No 14.27:

#### Question 1:

Find the coordinates of the point which divides the line segment joining (−1,3) and (4, −7) internally in the ratio 3 : 4

#### Answer:

We have A (−1, 3) and B (4,−7) be two points. Let a pointdivide the line segment joining the points A and B in the ratio 3:4 internally.

Now according to the section formula if point a point P divides a line segment joining andin the ratio m: n internally than,

Now we will use section formula to find the co-ordinates of unknown point P as,

Therefore, co-ordinates of point P is

#### Page No 14.27:

#### Question 2:

Find the points of trisection of the line segment joining the points:

(a) 5, −6 and (−7, 5),

(b) (3, −2) and (−3, −4),

(c) (2, −2) and (−7, 4).

#### Answer:

The co-ordinates of a point which divided two points and internally in the ratio is given by the formula,

The points of trisection of a line are the points which divide the line into the ratio.

(i) Here we are asked to find the points of trisection of the line segment joining the points *A*(5*,*−6) and *B*(−7*,*5).

So we need to find the points which divide the line joining these two points in the ratio and 2 : 1.

Let *P*(*x, y*) be the point which divides the line joining ‘*AB*’ in the ratio 1 : 2.

Let *Q*(*e, d*) be the point which divides the line joining ‘*AB*’ in the ratio 2 : 1.

Therefore the points of trisection of the line joining the given points are .

(ii) Here we are asked to find the points of trisection of the line segment joining the points *A*(3*,*−2) and *B*(−3*,*−4).

So we need to find the points which divide the line joining these two points in the ratio and 2 : 1.

Let *P*(*x, y*) be the point which divides the line joining ‘*AB*’ in the ratio 1 : 2.

Let *Q*(*e, d*) be the point which divides the line joining ‘*AB*’ in the ratio 2 : 1.

Therefore the points of trisection of the line joining the given points are.

(iii) Here we are asked to find the points of trisection of the line segment joining the points *A*(2*,*−2) and *B*(−7*,*4).

So we need to find the points which divide the line joining these two points in the ratio and 2 : 1.

Let *P*(*x, y*) be the point which divides the line joining ‘*AB*’ in the ratio 1 : 2.

Let *Q*(*e, d*) be the point which divides the line joining ‘*AB*’ in the ratio 2 : 1.

Therefore the points of trisection of the line joining the given points are .

#### Page No 14.27:

#### Question 3:

Find the coordinates of the point where the diagonals of the parallelogram formed by joining the points (−2, −1), (1, 0), (4, 3) and(1, 2) meet.

#### Answer:

The co-ordinates of the midpoint between two points and is given by,

In a parallelogram the diagonals bisect each other. That is the point of intersection of the diagonals is the midpoint of either of the diagonals.

Here, it is given that the vertices of a parallelogram are *A*(−2*,*−1)*, B*(1*,*0) and *C*(4*,*3) and *D*(1*,*2).

We see that ‘*AC*’ and ‘*BD*’ are the diagonals of the parallelogram.

The midpoint of either one of these diagonals will give us the point of intersection of the diagonals.

Let this point be *M*(*x, y*).

Let us find the midpoint of the diagonal ‘*AC*’.

Hence the co-ordinates of the point of intersection of the diagonals of the given parallelogram are.

#### Page No 14.27:

#### Question 4:

Prove that the points (3, −2), (4, 0), (6, −3) and (5, −5) are the vertices of a parallelogram.

#### Answer:

Let A (3,−2); B (4, 0); C (6,−3) and D (5,−5) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a parallelogram.

We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram.

Now to find the mid-point of two pointsand we use section formula as,

So the mid-point of the diagonal AC is,

Similarly mid-point of diagonal BD is,

Therefore the mid-points of the diagonals are coinciding and thus diagonal bisects each other.

Hence ABCD is a parallelogram.

#### Page No 14.27:

#### Question 5:

Three consecutive vertices of a parallelogram are (−2,−1), (1, 0) and (4, 3). Find the fourth vertex.

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (−2,−1); B (1, 0) and C (4, 3). We have to find the co-ordinates of the forth vertex.

Let the forth vertex be

Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.

Now to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

So the forth vertex is

#### Page No 14.27:

#### Question 6:

The points (3, −4) and (−6, 2) are the extremities of a diagonal of a parallelogram. If the third vertex is (−1,−3). Find the coordinates of the fourth vertex.

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (3,−4); B (−1,−3) and C (−6, 2). We have to find the co-ordinates of the forth vertex.

Let the forth vertex be

Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.

Now to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

So the forth vertex is

#### Page No 14.27:

#### Question 7:

Find the ratio in which the point (2, *y*) divides the line segment joining the points *A* (−2, 2) and *B* (3, 7). Also, find the value of *y*.

#### Answer:

The co-ordinates of a point which divided two points and internally in the ratio is given by the formula,

Here we are given that the point *P*(2*,y*) divides the line joining the points *A*(−2*,*2) and *B*(3*,*7) in some ratio.

Let us substitute these values in the earlier mentioned formula.

Equating the individual components we have

We see that the ratio in which the given point divides the line segment is.

Let us now use this ratio to find out the value of ‘*y*’.

Equating the individual components we have

Thus the value of ‘*y*’ is.

#### Page No 14.27:

#### Question 8:

If *A* (−1, 3), *B* (1, −1) and *C* (5, 1) are the vertices of a triangle *ABC*, find the length of the median through* A*.

#### Answer:

The distance *d* between two points and is given by the formula

The co-ordinates of the midpoint between two points and is given by,

Here, it is given that the three vertices of a triangle are *A*(−1*,*3)*, B*(1*,*−1) and *C*(5*,*1).

The median of a triangle is the line joining a vertex of a triangle to the mid-point of the side opposite this vertex.

Let ‘*D*’ be the mid-point of the side ‘*BC*’.

Let us now find its co-ordinates.

Thus we have the co-ordinates of the point as *D*(3*,*0).

Now, let us find the length of the median ‘*AD*’.

Thus the length of the median through the vertex ‘*A*’ of the given triangle is.

#### Page No 14.27:

#### Question 9:

If the coordinates of the mid-points of the sides of a triangle are (1, 1) (2, −3) and (3, 4), find the vertices of the triangle.

#### Answer:

The co-ordinates of the midpoint between two points and is given by,

Let the three vertices of the triangle be*,* and.

The three midpoints are given. Let these points be*, * and.

Let us now equate these points using the earlier mentioned formula,

Equating the individual components we get,

Using the midpoint of another side we have,

Equating the individual components we get,

Using the midpoint of the last side we have,

Equating the individual components we get,

Adding up all the three equations which have variable ‘*x*’ alone we have,

Substituting in the above equation we have,

Therefore,

And

Adding up all the three equations which have variable ‘*y*’ alone we have,

Substituting in the above equation we have,

Therefore,

And

Therefore the co-ordinates of the three vertices of the triangle are.

#### Page No 14.27:

#### Question 10:

If a vertex of a triangle be (1, 1) and the middle points of the sides through it be (−2,−3) and (5 2) find the other vertices.

#### Answer:

Let a in which P and Q are the mid-points of sides AB and AC respectively. The coordinates are: A (1, 1); P (−2, 3) and Q (5, 2).

We have to find the co-ordinates of and.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,

So, co-ordinates of B is (−5, 5)

Similarly, mid-point Q of side AC can be written as,

Now equate the individual terms to get,

So, co-ordinates of C is (9, 3)

#### Page No 14.27:

#### Question 11:

(i) In what ratio is the line segment joining the points (−2,−3) and (3, 7) divided by the y-axis? Also, find the coordinates of the point of division.

(ii) In what ratio is the line segment joining (−3, −1) and (−8, −9) divided at the point (−5, −21/5)?

#### Answer:

(i) The ratio in which the y-axis divides two points and is $\lambda :1$

The co-ordinates of the point dividing two points and in the ratio is given as,

; where

Here the two given points are *A*(−2*,*−3) and *B*(3*,*7).

Since, the point is on the y-axis so, *x* coordinate is 0.

$\frac{3\lambda -2}{1}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{2}{3}$

Thus the given points are divided by the y-axis in the ratio.

The co-ordinates of this point (*x, y*)* *can be found by using the earlier mentioned formula.

Thus the co-ordinates of the point which divides the given points in the required ratio are.

(ii) The co-ordinates of a point which divided two points and internally in the ratio is given by the formula,

Here it is said that the point divides the points (−3*,*−1) and (−8*,*−9). Substituting these values in the above formula we have,

Equating the individual components we have,

Therefore the ratio in which the line is divided is.

#### Page No 14.27:

#### Question 12:

If the mid-point of the line joining (3, 4) and (k, 7) is (x, y) and 2x + 2y + 1 = 0 find the value of *k*.

#### Answer:

We have two points A (3, 4) and B (k, 7) such that its mid-point is.

It is also given that point P lies on a line whose equation is

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,

Since, P lies on the given line. So,

Put the values of co-ordinates of point P in the equation of line to get,

On further simplification we get,

So,

#### Page No 14.27:

#### Question 13:

Determine the ratio in which the straight line x − y − 2 = 0 divides the line segment joining (3, −1) and (8, 9).

#### Answer:

Let the line divide the line segment joining the points A (3,−1) and B (8, 9) in the ratio at any point

Now according to the section formula if point a point P divides a line segment joining andin the ratio m: n internally than,

So,

Since, P lies on the given line. So,

Put the values of co-ordinates of point P in the equation of line to get,

On further simplification we get,

So,

So the line divides the line segment joining A and B in the ratio 2: 3 internally.

#### Page No 14.27:

#### Question 14:

Find the ratio in which the line segment joining (−2, −3) and (5, 6) is divided by (i) *x*-axis (ii) *y*-axis. Also, find the coordinates of the point of division in each case.

#### Answer:

The ratio in which the *x*−axis divides two points and is $\lambda :1$

The ratio in which the y-axis divides two points and is $\mu :1$

The co-ordinates of the point dividing two points and in the ratio is given as,

Where

Here the two given points are *A*(−2*,*−3) and *B*(5*,*6).

- The ratio in which the x-axis divides these points is

$\frac{6\lambda -3}{3}=0\phantom{\rule{0ex}{0ex}}\lambda =\frac{1}{2}$

Let point *P*(*x, y*) divide the line joining ‘*AB*’ in the ratio

Substituting these values in the earlier mentioned formula we have,

Thus the ratio in which the *x*−axis divides the two given points and the co-ordinates of the point is.

- The ratio in which the y-axis divides these points is

$\frac{5\mu -2}{3}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \mu =\frac{2}{5}$

Let point *P*(*x, y*) divide the line joining ‘*AB*’ in the ratio

Substituting these values in the earlier mentioned formula we have,

Thus the ratio in which the x-axis divides the two given points and the co-ordinates of the point is.

#### Page No 14.27:

#### Question 15:

Prove that the points (4, 5) (7, 6), (6, 3) (3, 2) are the vertices of a parallelogram. Is it a rectangle.

#### Answer:

Let A (4, 5); B (7, 6); C (6, 3) and D (3, 2) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a parallelogram.

We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram.

Now to find the mid-point of two pointsand we use section formula as,

So the mid-point of the diagonal AC is,

Similarly mid-point of diagonal BD is,

Therefore the mid-points of the diagonals are coinciding and thus diagonal bisects each other.

Hence ABCD is a parallelogram.

Now to check if ABCD is a rectangle, we should check the diagonal length.

Similarly,

Diagonals are of different lengths.

Hence ABCD is not a rectangle.

#### Page No 14.27:

#### Question 16:

Prove that (4, 3), (6, 4) (5, 6) and (3, 5) are the angular points of a square.

#### Answer:

Let A (4, 3); B (6, 4); C (5, 6) and D (3, 5) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a square.

So we should find the lengths of sides of quadrilateral ABCD.

All the sides of quadrilateral are equal.

So now we will check the lengths of the diagonals.

All the sides as well as the diagonals are equal. Hence ABCD is a square.

#### Page No 14.27:

#### Question 17:

Prove that the points (−4,−1), (−2, 4), (4, 0) and (2, 3) are the vertices of a rectangle.

#### Answer:

Let A (−4,−1); B (−2,−4); C (4, 0) and D (2, 3) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a rectangle.

So we should find the lengths of opposite sides of quadrilateral ABCD.

Opposite sides are equal. So now we will check the lengths of the diagonals.

Opposite sides are equal as well as the diagonals are equal. Hence ABCD is a rectangle.

#### Page No 14.27:

#### Question 18:

Find the lengths of the medians of a triangle whose vertices are A (−1,3), B(1,−1) and C(5, 1).

#### Answer:

We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (−1, 3); B (1,−1) and C (5, 1).

So we should find the mid-points of the sides of the triangle.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,

So co-ordinates of P is (0, 1)

Similarly mid-point Q of side BC can be written as,

Now equate the individual terms to get,

So co-ordinates of Q is (3, 0)

Similarly mid-point R of side AC can be written as,

Now equate the individual terms to get,

So co-ordinates of Q is (2, 2)

Therefore length of median from A to the side BC is,

Similarly length of median from B to the side AC is,

Similarly length of median from C to the side AB is

#### Page No 14.27:

#### Question 19:

Three vertices of a parallelogram are (a+b, a−b), (2a+b, 2a−b), (a−b, a+b). Find the fourth vertex.

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are;and. We have to find the co-ordinates of the forth vertex.

Let the forth vertex be

Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.

In general to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

So the forth vertex is

#### Page No 14.27:

#### Question 20:

If two vertices of a parallelogram are (3, 2) (−1, 0) and the diagonals cut at (2, −5), find the other vertices of the parallelogram.

#### Answer:

We have a parallelogram ABCD in which A (3, 2) and B (−1, 0) and the co-ordinate of the intersection of diagonals is M (2,−5).

We have to find the co-ordinates of vertices C and D.

So let the co-ordinates be and

In general to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

So the co-ordinate of vertex C is (1,−12)

Similarly,

Therefore,

Now equate the individual terms to get the unknown value. So,

So the co-ordinate of vertex C is (5,−10)

#### Page No 14.27:

#### Question 21:

If the coordinates of the mid-points of the sides of a triangle are (3, 4) (4, 6) and (5, 7), find its vertices.

#### Answer:

The co-ordinates of the midpoint between two points and is given by,

Let the three vertices of the triangle be*,* and.

The three midpoints are given. Let these points be*, * and.

Let us now equate these points using the earlier mentioned formula,

Equating the individual components we get,

Using the midpoint of another side we have,

Equating the individual components we get,

Using the midpoint of the last side we have,

Equating the individual components we get,

Adding up all the three equations which have variable ‘*x*’ alone we have,

Substituting in the above equation we have,

Therefore,

And

Adding up all the three equations which have variable ‘*y*’ alone we have,

Substituting in the above equation we have,

Therefore,

And

Therefore the co-ordinates of the three vertices of the triangle are.

#### Page No 14.28:

#### Question 22:

The line segment joining the points *P*(3, 3) and *Q*(6, −6) is trisected at the points *A* and *B* such that *A* is nearer to *P*. If A also lies on the line given by 2*x* + *y* + *k* = 0, find the value of* k*.

#### Answer:

We have two points P (3, 3) and Q (6,−6). There are two points A and B which trisect the line segment joining P and Q.

Let the co-ordinate of A be

Now according to the section formula if any point P divides a line segment joining andin the ratio m: n internally than,

The point A is the point of trisection of the line segment PQ. So, A divides PQ in the ratio 1: 2

Now we will use section formula to find the co-ordinates of unknown point A as,

Therefore, co-ordinates of point A is(4, 0)

It is given that point A lies on the line whose equation is

So point A will satisfy this equation.

So,

#### Page No 14.28:

#### Question 23:

If the points (−2, −1), (1, 0), (*x*, 3) and (1, *y*) form a parallelogram, find the values of *x* and *y*.

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (−2,−1); B (1, 0); C (*x*, 3) and D (1, *y*).

In general to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

Similarly,

Therefore,

#### Page No 14.28:

#### Question 24:

The points A(2, 0), B(9, 1) C(11, 6) and D(4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.

#### Answer:

Let A (2, 0); B (9, 1); C (11, 6) and D (4, 4) be the vertices of a quadrilateral. We have to check if the quadrilateral ABCD is a rhombus or not.

So we should find the lengths of sides of quadrilateral ABCD.

All the sides of quadrilateral are unequal. Hence ABCD is not a rhombus.

#### Page No 14.28:

#### Question 25:

If three consecutive vertices of a parallelogram are (1, −2), (3, 6) and (5, 10), find its fourth vertex.

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (1,−2);

B (3, 6) and C(5, 10). We have to find the co-ordinates of the forth vertex.

Let the forth vertex be

Now to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

Similarly,

So the forth vertex is

#### Page No 14.28:

#### Question 26:

If the points A (a, −11), B (5, b), C(2, 15) and D (1, 1) are the vertices of a parallelogram ABCD, find the values of *a* and *b*.

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (*a*,−11); B (5, *b*); C (2, 15) and D (1, 1).

In general to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

Similarly,

Therefore,

#### Page No 14.28:

#### Question 27:

If the coordinates of the mid-points of the sides of a triangle be (3, −2), (−3, 1) and (4, −3), then find the coordinates of its vertices.

#### Answer:

The co-ordinates of the midpoint between two points and is given by,

Let the three vertices of the triangle be*,* and.

The three midpoints are given. Let these points be*, * and.

Let us now equate these points using the earlier mentioned formula,

Equating the individual components we get,

Using the midpoint of another side we have,

Equating the individual components we get,

Using the midpoint of the last side we have,

Equating the individual components we get,

Adding up all the three equations which have variable ‘*x*’ alone we have,

Substituting in the above equation we have,

Therefore,

And

Adding up all the three equations which have variable ‘*y*’ alone we have,

Substituting in the above equation we have,

Therefore,

And

Therefore the co-ordinates of the three vertices of the triangle are.

#### Page No 14.28:

#### Question 28:

Find the length of the medians of a Δ*ABC* having vertices at A(0, −1), B(2, 1) and C(0, 3).

#### Answer:

We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (0,−1); B (2, 1) and C (0, 3).

So we should find the mid-points of the sides of the triangle.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,

So co-ordinates of P is (1, 0)

Similarly mid-point Q of side BC can be written as,

Now equate the individual terms to get,

So co-ordinates of Q is (1, 2)

Similarly mid-point R of side AC can be written as,

Now equate the individual terms to get,

So co-ordinates of R is (0, 1)

Therefore length of median from A to the side BC is,

Similarly length of median from B to the side AC is,

Similarly length of median from C to the side AB is

#### Page No 14.28:

#### Question 29:

Find the lengths of the medians of a Δ*ABC* having vertices at *A*(5, 1), B(1, 5), and *C*(−3, −1).

#### Answer:

We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (5, 1); B (1, 5) and C (−3,−1).

So we should find the mid-points of the sides of the triangle.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,

So co-ordinates of P is (3, 3)

Similarly mid-point Q of side BC can be written as,

Now equate the individual terms to get,

So co-ordinates of Q is (−1, 2)

Similarly mid-point R of side AC can be written as,

Now equate the individual terms to get,

So co-ordinates of R is (1, 0)

Therefore length of median from A to the side BC is,

Similarly length of median from B to the side AC is,

Similarly length of median from C to the side AB is

#### Page No 14.28:

#### Question 30:

Find the coordinates of the points which divide the line segment joining the points (−4, 0) and (0, 6) in four equal parts.

#### Answer:

The co-ordinates of the midpoint between two points and is given by,

Here we are supposed to find the points which divide the line joining *A*(−4*,*0) and *B*(0*,*6) into 4 equal parts.

We shall first find the midpoint *M*(*x, y*)* *of these two points since this point will divide the line into two equal parts

So the point *M*(−2*,*3) splits this line into two equal parts.

Now, we need to find the midpoint of *A*(−4*,*0) and *M*(−2*,*3) separately and the midpoint of *B*(0*,*6) and *M*(−2*,*3). These two points along with *M*(−2*,*3) split the line joining the original two points into four equal parts.

Let be the midpoint of *A*(−4*,*0) and *M*(−2*,*3).

Now let bet the midpoint of *B*(0*,*6) and *M*(−2*,*3).

Hence the co-ordinates of the points which divide the line joining the two given points are.

#### Page No 14.28:

#### Question 31:

Show that the mid-point of the line segment joining the points (5, 7) and (3, 9) is also the mid-point of the line segment joining the points (8, 6) and (0, 10).

#### Answer:

We have two points A (5, 7) and B (3, 9) which form a line segment and similarly

C (8, 6) and D (0, 10) form another line segment.

We have to prove that mid-point of AB is also the mid-point of CD.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of line segment AB can be written as,

Now equate the individual terms to get,

So co-ordinates of P is (4, 8)

Similarly mid-point Q of side CD can be written as,

Now equate the individual terms to get,

So co-ordinates of Q is (4, 8)

Hence the point P and Q coincides.

Thus mid-point of AB is also the mid-point of CD.

#### Page No 14.28:

#### Question 32:

Find the distance of the point (1, 2) from the mid-point of the line segment joining the points (6, 8) and (2, 4).

#### Answer:

We have to find the distance of a point A (1, 2) from the mid-point of the line segment joining P (6, 8) and Q (2, 4).

In general to find the mid-point of any two pointsand we use section formula as,

Therefore mid-point B of line segment PQ can be written as,

Now equate the individual terms to get,

So co-ordinates of B is (4, 6)

Therefore distance between A and B,

#### Page No 14.28:

#### Question 33:

If *A* and *B* are (1, 4) and (5, 2) respectively, find the coordinates of *P* when *AP*/*BP* = 3/4.

#### Answer:

The co-ordinates of the point dividing two points and in the ratio is given as,

where,

Here the two given points are *A*(1*,*4) and *B*(5*,*2). Let point *P*(*x, y*) divide the line joining ‘*AB*’ in the ratio

Substituting these values in the earlier mentioned formula we have,

Thus the co-ordinates of the point which divides the given points in the required ratio are.

#### Page No 14.28:

#### Question 34:

Show that the points *A* (1, 0), *B* (5, 3), *C* (2, 7) and *D* (−2, 4) are the vertices of a parallelogram.

#### Answer:

Let A (1, 0); B (5, 3); C (2, 7) and D (−2, 4) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a parallelogram.

We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram.

Now to find the mid-point of two pointsand we use section formula as,

So the mid-point of the diagonal AC is,

Similarly mid-point of diagonal BD is,

Therefore the mid-points of the diagonals are coinciding and thus diagonal bisects each other.

Hence ABCD is a parallelogram.

#### Page No 14.28:

#### Question 35:

Determine the ratio in which the point P (m, 6) divides the join of *A*(−4, 3) and *B*(2, 8). Also, find the value of *m*.

#### Answer:

The co-ordinates of a point which divided two points and internally in the ratio is given by the formula,

Here we are given that the point *P*(*m,*6) divides the line joining the points *A*(−4*,*3) and *B*(2*,*8) in some ratio.

Let us substitute these values in the earlier mentioned formula.

Equating the individual components we have

We see that the ratio in which the given point divides the line segment is.

Let us now use this ratio to find out the value of ‘*m*’.

Equating the individual components we have

Thus the value of ‘*m*’ is.

#### Page No 14.28:

#### Question 36:

Determine the ratio in which the point (−6, *a*) divides the join of *A* (−3, 1) and *B* (−8, 9). Also find the value of *a*.

#### Answer:

Here we are given that the point *P*(−6*,a*) divides the line joining the points *A*(−3*,*1) and *B*(−8*,*9) in some ratio.

Let us substitute these values in the earlier mentioned formula.

Equating the individual components we have

We see that the ratio in which the given point divides the line segment is.

Let us now use this ratio to find out the value of ‘*a*’.

Equating the individual components we have

Thus the value of ‘*a*’ is.

#### Page No 14.28:

#### Question 37:

The line segment joining the points (3, −4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, −2) and (5/3, q) respectively. Find the values of* p* and *q*.

#### Answer:

We have two points A (3,−4) and B (1, 2). There are two points P (*p*,−2) and Q which trisect the line segment joining A and B.

Now according to the section formula if any point P divides a line segment joining andin the ratio m: n internally than,

The point P is the point of trisection of the line segment AB. So, P divides AB in the ratio 1: 2

Now we will use section formula to find the co-ordinates of unknown point A as,

Equate the individual terms on both the sides. We get,

Similarly, the point Q is the point of trisection of the line segment AB. So, Q divides AB in the ratio 2: 1

Now we will use section formula to find the co-ordinates of unknown point A as,

Equate the individual terms on both the sides. We get,

#### Page No 14.28:

#### Question 38:

The line joining the points (2, 1) and (5, −8) is trisected at the points *P* and *Q*. If point *P* lies on the line 2*x* − *y* + *k* = 0. Find the value of k.

#### Answer:

We have two points A (2, 1) and B (5,−8). There are two points P and Q which trisect the line segment joining A and B.

Now according to the section formula if any point P divides a line segment joining andin the ratio m: n internally than,

The point P is the point of trisection of the line segment AB. So, P divides AB in the ratio 1: 2

Now we will use section formula to find the co-ordinates of unknown point A as,

Therefore, co-ordinates of point P is(3,−2)

It is given that point P lies on the line whose equation is

So point A will satisfy this equation.

So,

#### Page No 14.28:

#### Question 39:

If *A* and *B* are two points having coordinates (−2, −2) and (2, −4) respectively, find the coordinates of *P* such that *AP* = $\frac{3}{7}$*AB.*

#### Answer:

We have two points A (−2,−2) and B (2,−4). Let P be any point which divide AB as,

Since,

So,

Therefore P divides AB in the ratio 3: 4. So,

#### Page No 14.28:

#### Question 40:

Find the coordinates of the points which divide the line segment joining A(−2, 2) and B (2, 8) into four equal parts.

#### Answer:

The co-ordinates of the midpoint between two points and is given by,

Here we are supposed to find the points which divide the line joining *A*(−2*,*2) and *B*(2*,*8) into 4 equal parts.

We shall first find the midpoint *M*(*x, y*)* *of these two points since this point will divide the line into two equal parts.

So the point *M*(0*,*5) splits this line into two equal parts.

Now, we need to find the midpoint of *A*(−2*,*2) and *M*(0*,*5) separately and the midpoint of *B*(2*,*8) and *M*(0*,*5). These two points along with *M*(0*,*5) split the line joining the original two points into four equal parts.

Let be the midpoint of *A*(−2*,*2) and *M*(0*,*5).

Now let bet the midpoint of *B*(2*,*8) and *M*(0*,*5).

Hence the co-ordinates of the points which divide the line joining the two given points are.

#### Page No 14.28:

#### Question 41:

*A* (4, 2), *B*(6, 5) and *C* (1, 4) are the vertices of Δ*ABC*.

(i) The median from *A* meets *BC* in *D.* Find the coordinates of the point *D*.

(ii) Find the coordinates of point *P* and *AD* such that *AP* : *PD* = 2 : 1.

(iii) Find the coordinates of the points* Q* and *R* on medians *BE* and *CF *respectively such that *BQ* : *QE* = 2 : 1 and *CR* : *RF* = 2 : 1

(iv) What do you observe?

#### Answer:

We have triangle in which the co-ordinates of the vertices are A (4, 2); B (6, 5) and C (1, 4)

(i)It is given that median from vertex A meets BC at D. So, D is the mid-point of side BC.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point D of side BC can be written as,

Now equate the individual terms to get,

So co-ordinates of D is

(ii)We have to find the co-ordinates of a point P which divides AD in the ratio 2: 1 internally.

P divides AD in the ratio 2: 1. So,

(iii)We need to find the mid-point of sides AB and AC. Let the mid-points be F and E for the sides AB and AC respectively.

Therefore mid-point F of side AB can be written as,

So co-ordinates of F is

Similarly mid-point E of side AC can be written as,

So co-ordinates of E is

Q divides BE in the ratio 2: 1. So,

Similarly, R divides CF in the ratio 2: 1. So,

(iv)We observe that that the point P, Q and R coincides with the centroid. This also shows that centroid divides the median in the ratio 2: 1.

#### Page No 14.29:

#### Question 48:

A point P divides the line segment joining the points A(3, −5) and B(−4, 8) such that $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{k}{1}.$ If P lies on the line *x* + *y* = 0, then find the value of *k*. [CBSE 2012]

#### Answer:

It is given that $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{k}{1}.$

So, P divides the line segment joining the points A(3, −5) and B(−4, 8) in the ratio *k* : 1.

Using the section formula, we get

Coordinates of P = $\left(\frac{-4k+3}{k+1},\frac{8k-5}{k+1}\right)$

Since P lies on the line *x* + *y* = 0, so

$\frac{-4k+3}{k+1}+\frac{8k-5}{k+1}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{-4k+3+8k-5}{k+1}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4k-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{1}{2}$

Hence, the value of *k* is $\frac{1}{2}$.

#### Page No 14.29:

#### Question 49:

Find the ratio in which the point P(−1, *y*) lying on the line segment joining A(−3, 10) and B(6 −8) divides it. Also find the value of *y*. [CBSE 2013]

#### Answer:

Suppose P(−1, *y*) divides the line segment joining A(−3, 10) and B(6 −8) in the ratio *k* : 1.

Using section formula, we get

Coordinates of P = $\left(\frac{6k-3}{k+1},\frac{-8k+10}{k+1}\right)$

$\therefore \left(\frac{6k-3}{k+1},\frac{-8k+10}{k+1}\right)=\left(-1,y\right)$

$\Rightarrow \frac{6k-3}{k+1}=-1$ and $y=\frac{-8k+10}{k+1}$

Now,

$\frac{6k-3}{k+1}=-1\phantom{\rule{0ex}{0ex}}\Rightarrow 6k-3=-k-1\phantom{\rule{0ex}{0ex}}\Rightarrow 7k=2\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{2}{7}$

So, P divides the line segment AB in the ratio 2 : 7.

Putting *k* = $\frac{2}{7}$ in $y=\frac{-8k+10}{k+1}$, we get

$y=\frac{-8\times {\displaystyle \frac{2}{7}}+10}{{\displaystyle \frac{2}{7}}+1}=\frac{-16+70}{2+7}=\frac{54}{9}=6$

Hence, the value of *y* is 6.

#### Page No 14.29:

#### Question 42:

*ABCD* is a rectangle formed by joining the points *A* (−1, −1), *B*(−1 4) *C* (5 4) and *D* (5, −1). *P*, *Q*, *R* and *S* are the mid-points of sides *AB*, *BC*, *CD* and *DA* respectively. Is the quadrilateral *PQRS* a square? a rectangle? or a rhombus? Justify your answer.

#### Answer:

We have a rectangle ABCD formed by joining the points A (−1,−1); B (−1, 4); C (5, 4) and D (5,−1). The mid-points of the sides AB, BC, CD and DA are P, Q, R, S respectively.

We have to find that whether PQRS is a square, rectangle or rhombus.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,

So co-ordinates of P is

Similarly mid-point Q of side BC can be written as,

Now equate the individual terms to get,

So co-ordinates of Q is (2, 4)

Similarly mid-point R of side CD can be written as,

Now equate the individual terms to get,

So co-ordinates of R is

Similarly mid-point S of side DA can be written as,

Now equate the individual terms to get,

So co-ordinates of S is (2,−1)

So we should find the lengths of sides of quadrilateral PQRS.

All the sides of quadrilateral are equal.

So now we will check the lengths of the diagonals.

All the sides are equal but the diagonals are unequal. Hence ABCD is a rhombus.

#### Page No 14.29:

#### Question 43:

Show that *A* (−3, 2), *B* (−5, −5), *C *(2,−3), and *D* (4, 4) are the vertices of a rhombus.

#### Answer:

Let A (−3, 2); B (−5,−5); C (2,−3) and D (4, 4) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a rhombus.

So we should find the lengths of sides of quadrilateral ABCD.

All the sides of quadrilateral are equal. Hence ABCD is a rhombus.

#### Page No 14.29:

#### Question 44:

Find the ratio in which the *y*-axis divides the line segment joining the points (5, −6) and (−1,−4). Also, find the coordinates of the point of division.

#### Answer:

The ratio in which the y-axis divides two points and is $\lambda :1$

The co-ordinates of the point dividing two points and in the ratio is given as,

where,

Here the two given points are *A*(5*,*−6) and *B*(−1*,*−4).

$(x,y)=\left(\frac{-\lambda +5}{\lambda +1},\frac{-4\lambda -6}{\lambda +1}\right)$

Since, the y-axis divided the given line, so the x coordinate will be 0.

$\frac{-\lambda +5}{\lambda +1}=0\phantom{\rule{0ex}{0ex}}\lambda =\frac{5}{1}$

Thus the given points are divided by the y-axis in the ratio.

The co-ordinates of this point (*x, y*)* *can be found by using the earlier mentioned formula.

Thus the co-ordinates of the point which divides the given points in the required ratio are.

#### Page No 14.29:

#### Question 45:

If the points *A* (6, 1), *B* (8, 2), *C* (9, 4) and *D* (*k*, *p*) are the vertices of a parallelogram taken in order, then find the values of *k* and *p*.

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (6, 1); B (8, 2); C (9, 4) and D (*k, p*).

In general to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

Similarly,

Therefore,

#### Page No 14.29:

#### Question 46:

In what ratio does the point (−4, 6) divide the line segment joining the points A(−6, 10) and B(3,−8)?

#### Answer:

Here it is said that the point (−4*,*6)* *divides the points *A*(−6*,*10) and *B*(3*,*−8). Substituting these values in the above formula we have,

Equating the individual components we have,

Therefore the ratio in which the line is divided is

#### Page No 14.29:

#### Question 47:

Find the coordinates of a point *A*, where *AB* is a diameter of the circle whose centre is (2, −3) and *B* is (1, 4).

#### Answer:

Let the co-ordinates of point A be.

Centre lies on the mid-point of the diameter. So applying the mid-point formula we get,

Similarly,

So the co-ordinates of A are (3,−10)

#### Page No 14.35:

#### Question 6:

If *G* be the centroid of a triangle ABC, prove that:

AB^{2} + BC^{2} + CA^{2} = 3 (GA^{2} + GB^{2} + GC^{2})

#### Answer:

$\mathrm{Let}\mathrm{A}\left({x}_{1},{y}_{1}\right);\mathrm{B}\left({x}_{2},{y}_{2}\right);\mathrm{C}\left({x}_{3},{y}_{3}\right)\mathrm{be}\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{the}\mathrm{vertices}\mathrm{of}\u2206\mathrm{ABC}.\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{us}\mathrm{assume}\mathrm{that}\mathrm{centroid}\mathrm{of}\mathrm{the}\u2206\mathrm{ABC}\mathrm{is}\mathrm{at}\mathrm{the}\mathrm{origin}\mathrm{G}.\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{G}\mathrm{are}\mathrm{G}\left(0,0\right).\phantom{\rule{0ex}{0ex}}\mathrm{Now},\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3}=0;\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}=0\phantom{\rule{0ex}{0ex}}\mathrm{so},{x}_{1}+{x}_{2}+{x}_{3}=0.......\left(1\right)\phantom{\rule{0ex}{0ex}}{y}_{1}+{y}_{2}+{y}_{3}=0........\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{Squaring}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}+2{x}_{1}{x}_{2}+2{x}_{2}{x}_{3}+2{x}_{3}{x}_{1}=0.......\left(3\right)\phantom{\rule{0ex}{0ex}}{{y}_{1}}^{2}+{{y}_{2}}^{2}+{{y}_{3}}^{2}+2{y}_{1}{y}_{2}+2{y}_{2}{y}_{3}+2{y}_{3}{y}_{1}=0......\left(4\right)\phantom{\rule{0ex}{0ex}}\mathrm{LHS}={\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}+{\mathrm{CA}}^{2}\phantom{\rule{0ex}{0ex}}={\left[\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\right]}^{2}+{\left[\sqrt{{\left({x}_{3}-{x}_{2}\right)}^{2}+{\left({y}_{3}-{y}_{2}\right)}^{2}}\right]}^{2}+{\left[\sqrt{{\left({x}_{3}-{x}_{1}\right)}^{2}+{\left({y}_{3}-{y}_{1}\right)}^{2}}\right]}^{2}\phantom{\rule{0ex}{0ex}}={\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}+{\left({x}_{3}-{x}_{2}\right)}^{2}+{\left({y}_{3}-{y}_{2}\right)}^{2}+{\left({x}_{3}-{x}_{1}\right)}^{2}+{\left({y}_{3}-{y}_{1}\right)}^{2}\phantom{\rule{0ex}{0ex}}={{x}_{1}}^{2}+{{x}_{2}}^{2}-2{x}_{1}{x}_{2}+{{y}_{1}}^{2}+{{y}_{2}}^{2}-2{y}_{1}{y}_{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}-2{x}_{2}{x}_{3}+{{y}_{2}}^{2}+{{y}_{3}}^{2}-2{y}_{2}{y}_{3}+{{x}_{1}}^{2}+{{x}_{3}}^{2}-2{x}_{1}{x}_{3}+{{y}_{1}}^{2}+{{y}_{3}}^{2}-2{y}_{1}{y}_{3}\phantom{\rule{0ex}{0ex}}=2\left({{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}\right)+2\left({{y}_{1}}^{2}+{{y}_{2}}^{2}+{{y}_{3}}^{2}\right)-\left(2{x}_{1}{x}_{2}+2{x}_{2}{x}_{3}+2{x}_{3}{x}_{1}\right)-\left(2{y}_{1}{y}_{2}+2{y}_{2}{y}_{3}+2{y}_{3}{y}_{1}\right)\phantom{\rule{0ex}{0ex}}=2\left({{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}\right)+2\left({{y}_{1}}^{2}+{{y}_{2}}^{2}+{{y}_{3}}^{2}\right)+\left({{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}\right)+\left({{y}_{1}}^{2}+{{y}_{2}}^{2}+{{y}_{3}}^{2}\right)\phantom{\rule{0ex}{0ex}}=3\left({{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}+{{y}_{1}}^{2}+{{y}_{2}}^{2}+{{y}_{3}}^{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{RHS}=3\left({\mathrm{GA}}^{2}+{\mathrm{GB}}^{2}+{\mathrm{GC}}^{2}\right)\phantom{\rule{0ex}{0ex}}=3\left[{\left\{\sqrt{{\left({x}_{1}-0\right)}^{2}+{\left({y}_{1}-0\right)}^{2}}\right\}}^{2}+{\left\{\sqrt{{\left({x}_{2}-0\right)}^{2}+{\left({y}_{2}-0\right)}^{2}}\right\}}^{2}+{\left\{\sqrt{{\left({x}_{3}-0\right)}^{2}+{\left({y}_{3}-0\right)}^{2}}\right\}}^{2}\right]\phantom{\rule{0ex}{0ex}}=3\left({{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}+{{y}_{1}}^{2}+{{y}_{2}}^{2}+{{y}_{3}}^{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Hence},{\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}+{\mathrm{CA}}^{2}=3\left({\mathrm{GA}}^{2}+{\mathrm{GB}}^{2}+{\mathrm{GC}}^{2}\right)$

#### Page No 14.35:

#### Question 1:

Find the centroid of the triangle whose vertices are:

(i) (1, 4) (−1,−1), (3, −2)

(ii) (−2, 3) (2, −1) (4, 0)

#### Answer:

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

(i) The co-ordinates of the centroid of a triangle whose vertices are (1, 4); (−1,−1); (3,−2) are-

(ii) The co-ordinates of the centroid of a triangle whose vertices are (−2, 3); (2,−1); (4, 0) are-

#### Page No 14.35:

#### Question 2:

Two vertices of a triangle are (1, 2), (3, 5) and its centroid is at the origin. Find the coordinates of the third vertex.

#### Answer:

We have to find the co-ordinates of the third vertex of the given triangle. Let the co-ordinates of the third vertex be.

The co-ordinates of other two vertices are (1, 2) and (3, 5)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are is−

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the co-ordinate of third vertex

#### Page No 14.35:

#### Question 3:

Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.

#### Answer:

Letbe any triangle such that O is the origin and the other co-ordinates are. P and R are the mid-points of the sides OA and OB respectively.

We have to prove that line joining the mid-point of any two sides of a triangle is equal to half of the third side which means,

In general to find the mid-point of two pointsand we use section formula as,

So,

Co-ordinates of P is,

Similarly, co-ordinates of R is,

In general, the distance between A and B is given by,

Similarly,

Hence,

#### Page No 14.35:

#### Question 4:

Prove that the lines joining the middle points of the opposite sides of a quadrilateral and the join of the middle points of its diagonals meet in a point and bisect one another.

#### Answer:

Let us consider a Cartesian plane having a parallelogram OABC in which O is the origin.

We have to prove that middle point of the opposite sides of a quadrilateral and the join of the mid-points of its diagonals meet in a point and bisect each other.

Let the co-ordinate of A be. So the coordinates of other vertices of the quadrilateral are- O (0, 0); B; C

Let P, Q, R and S be the mid-points of the sides AB, BC, CD, DA respectively.

In general to find the mid-point of two pointsand we use section formula as,

So co-ordinate of point P,

Similarly co-ordinate of point Q,

Similarly co-ordinate of point R,

Similarly co-ordinate of point S,

Let us find the co-ordinates of mid-point of PR as,

Similarly co-ordinates of mid-point of QS as,

Now the mid-point of diagonal AC,

Similarly the mid−point of diagonal OA,

Hence the mid-points of PR, QS, AC and OA coincide.

Thus, middle point of the opposite sides of a quadrilateral and the join of the mid-points of its diagonals meet in a point and bisect each other.

#### Page No 14.35:

#### Question 5:

If G be the centroid of a triangle ABC and P be any other point in the plane, prove that PA^{2} + PB^{2} + PC^{2} = GA^{2} + GB^{2} + GC^{2} + 3GP^{2}.

#### Answer:

Letbe any triangle whose coordinates are. Let P be the origin and G be the centroid of the triangle.

We have to prove that,

…… (1)

We know that the co-ordinates of the centroid G of a triangle whose vertices are is−

In general, the distance between A and B is given by,

So,

Now,

So we get the value of left hand side of equation (1) as,

Similarly we get the value of right hand side of equation (1) as,

Hence,

#### Page No 14.35:

#### Question 7:

If (−2, 3), (4, −3) and (4, 5) are the mid-points of the sides of a triangle, find the coordinates of its centroid.

#### Answer:

Letbe ant triangle such that P (−2, 3); Q (4,−3) and R (4, 5) are the mid-points of the sides AB, BC, CA respectively.

We have to find the co-ordinates of the centroid of the triangle.

Let the vertices of the triangle be

In general to find the mid-point of two pointsand we use section formula as,

So, co-ordinates of P,

Equate the *x* component on both the sides to get,

…… (1)

Similarly,

…… (2)

Similarly, co-ordinates of Q,

Equate the *x* component on both the sides to get,

…… (3)

Similarly,

…… (4)

Similarly, co-ordinates of R,

Equate the *x* component on both the sides to get,

…… (5)

Similarly,

…… (6)

Add equation (1) (3) and (5) to get,

Similarly, add equation (2) (4) and (6) to get,

We know that the co-ordinates of the centroid G of a triangle whose vertices are is-

So, centroid G of a triangle is,

#### Page No 14.35:

#### Question 8:

In Fig. 14.36, a right triangle BOA is given C is the mid-point of the hypotenuse AB. Show that it is equidistant from the vertices O, A and B.

#### Answer:

We have a right angled triangle, right angled at O. Co-ordinates are B (0,2*b*); A (2*a**, *0) and C (0, 0).

We have to prove that mid-point C of hypotenuse AB is equidistant from the vertices.

In general to find the mid-point of two pointsand we use section formula as,

So co-ordinates of C is,

In general, the distance between A and B is given by,

So,

Hence, mid−point C of hypotenuse AB is equidistant from the vertices.

#### Page No 14.35:

#### Question 9:

Find the third vertex of a triangle, if two of its vertices are at (−3, 1) and (0, −2) and the centroid is at the origin.

#### Answer:

We have to find the co-ordinates of the third vertex of the given triangle. Let the co-ordinates of the third vertex be.

The co-ordinates of other two vertices are (−3, 1) and (0, −2)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the co-ordinate of third vertex

#### Page No 14.35:

#### Question 10:

*A* (3, 2) and *B* (−2, 1) are two vertices of a triangle *ABC *whose centroid *G* has the coordinates $\left(\frac{5}{3},-\frac{1}{3}\right)$. Find the coordinates of the third vertex *C* of the triangle.

#### Answer:

We have to find the co-ordinates of the third vertex of the given triangle. Let the co-ordinates of the third vertex be.

The co-ordinates of other two vertices are A (3, 2) and C (−2, 1)

The co-ordinate of the centroid is

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the co-ordinate of third vertex

#### Page No 14.4:

#### Question 1:

On which axis do the following points lie?

(a) P(5, 0)

(b) Q(0−2)

(c) R(−4,0)

(d) S(0,5)

#### Answer:

According to the Rectangular Cartesian Co-ordinate system of representing a point *(x, y)*,

If then the point lies in the 1^{st} quadrant

If then the point lies in the 2^{nd} quadrant

If then the point lies in the 3^{rd} quadrant

If then the point lies in the 4^{th} quadrant

But in case

If then the point lies on the *y*-axis

If then the point lies on the *x*-axis

(i) Here the point is given to be *P *(5, 0). Comparing this with the standard form of

(*x, y*) we have

Here we see that

Hence the given point lies on the

(ii) Here the point is given to be *Q* *(0, *-−2). Comparing this with the standard form of *(x, y)* we have

Here we see that

Hence the given point lies on the

(iii) Here the point is given to be *R *(-4, 0). Comparing this with the standard form of (*x, y*) we have

Here we see that

Hence the given point lies on the

(iv) Here the point is given to be S (0, 5). Comparing this with the standard form of (*x, y*) we have

Here we see that

Hence the given point lies on the

#### Page No 14.4:

#### Question 2:

Let *ABCD* be a square of side 2*a*. Find the coordinates of the vertices of this square when

(i) A coincides with the origin and *AB* and *AD* are along *OX* and *OY* respectively.

(ii) The centre of the square is at the origin and coordinate axes are parallel to the sides *AB* and *AD* respectively.

#### Answer:

The distance between any two adjacent vertices of a square will always be equal. This distance is nothing but the side of the square.

Here, the side of the square ‘*ABCD*’ is given to be ‘2*a*’.

(i) Since it is given that the vertex ‘*A*’ coincides with the origin we know that the co-ordinates of this point is (0, 0).

We also understand that the side ‘*AB*’ is along the *x*-axis. So, the vertex ‘*B*’ has got to be at a distance of ‘2*a*’ from ‘*A*’.

Hence the vertex ‘*B*’ has the co-ordinates (2*a**, *0).

Also it is said that the side ‘*AD*’ is along the *y*-axis. So, the vertex ‘*D*’ it has got to be at a distance of ‘*2a*’ from ‘*A*’.

Hence the vertex ‘*D*’ has the co-ordinates (0, 2*a*)

Finally we have vertex ‘*C*’ at a distance of ‘2*a*’ both from vertex ‘*B*’ as well as ‘*D*’.

Hence the vertex of ‘*C*’ has the co-ordinates (2*a**, *2*a*)

So, the co-ordinates of the different vertices of the square are

(ii) Here it is said that the centre of the square is at the origin and that the sides of the square are parallel to the axes.

Moving a distance of half the side of the square in either the ‘*upward*’ or ‘*downward*’ direction and also along either the ‘*right*’ or ‘*left*’ direction will give us all the four vertices of the square.

Half the side of the given square is ‘*a*’.

The centre of the square is the origin and its vertices are (0, 0).* *Moving a distance of ‘*a*’ to the right as well as up will lead us to the vertex ‘*A*’ and it will have vertices (*a, a*).

Moving a distance of ‘*a*’ to the left as well as up will lead us to the vertex ‘*B*’ and it will have vertices (-(−*a**, a*).

Moving a distance of ‘*a*’ to the left as well as down will lead us to the vertex ‘*C*’ and it will have vertices (-(−*a**, *-−*a*).

Moving a distance of ‘*a*’ to the right as well as down will lead us to the vertex ‘*D*’ and it will have vertices (*a,-,−a*).

So, the co-ordinates of the different vertices of the square are

#### Page No 14.4:

#### Question 3:

The base *PQ* of two equilateral triangles *PQR* and *PQR'* with side 2*a* lies along y-axis such that the mid-point of *PQ* is at the origin. Find the coordinates of the vertices *R* and* R*' of the triangles.

#### Answer:

In an equilateral triangle, the height ‘*h*’ is given by

Here it is given that ‘*PQ*’ forms the base of two equilateral triangles whose side measures ‘*2a*’ units.

The height of these two equilateral triangles has got to be

In an equilateral triangle the height drawn from one vertex meets the midpoint of the side opposite this vertex.

So here we have ‘*PQ*’ being the base lying along the *y*-axis with its midpoint at the origin, that is at *(0, 0)*.

So the vertices ‘*R*’ and ‘*R’*’ will lie perpendicularly to the *y*-axis on either sides of the origin at a distance of ‘’ units.

Hence the co-ordinates of ‘*R*’ and ‘*R’*’ are

#### Page No 14.47:

#### Question 1:

Find the area of a triangle whose vertices are

(i) (6, 3) (−3, 5) and (4, −2)

(ii) $(a{t}_{1}^{2},2a{t}_{1}),(a{t}_{2}^{2},2a{t}_{2})and(a{t}_{3}^{2},2a{t}_{3})$

(iii) (a, c + a), (a, c) and (−a, c − a)

#### Answer:

We know area of triangle formed by three points is given by

(i) The vertices are given as (6, 3), (−3, 5), (4, −2).

(ii) The vertices are given as .

(iii)

The vertices are given as .

#### Page No 14.47:

#### Question 2:

Find the area of the quadrilaterals, the coordinates of whose vertices are

(i) (−3, 2), (5, 4), (7, −6) and (−5, −4)

(ii) (1, 2), (6, 2), (5, 3) and (3, 4)

(iii) (−4, −2, (−3, −5), (3, −2), (2, 3)

#### Answer:

(i)

Let the vertices of the quadrilateral be A (−3, 2), B (5, 4), C (7, −6), and D (−5, −4). Join AC to form two triangles ΔABC and ΔACD.

(ii)

Let the vertices of the quadrilateral be A (1, 2), B (6, 2), C (5, 3), and D (3, 4). Join AC to form two triangles ΔABC and ΔACD.

(iii)

Let the vertices of the quadrilateral be A (−4, −2), B (−3, −5), C (3, −2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.

#### Page No 14.48:

#### Question 3:

The four vertices of a quadrilateral are (1, 2), (−5, 6), (7, −4) and (k, −2) taken in order. If the area of the quadrilateral is zero, find the value of *k*.

#### Answer:

GIVEN: The four vertices of quadrilateral are (1, 2), (−5, 6), (7, −4) and D (*k*, −2) taken in order. If the area of the quadrilateral is zero

TO FIND: value of *k*

PROOF: Let four vertices of quadrilateral are A (1, 2) and B (−5, 6) and C (7, −4) and D (*k*, −2)

We know area of triangle formed by three points is given by

Now Area of ΔABC

Taking three points when A (1, 2) and B (−5, 6) and C (7, −4)

Also,

Now Area of ΔACD

Taking three points when A (1, 2) and C (7, −4) and D (*k*, −2)

Hence

#### Page No 14.48:

#### Question 4:

The vertices of Δ*ABC* are (−2, 1), (5, 4) and (2, −3) respectively. Find the area of the triangle and the length of the altitude through *A*.

#### Answer:

GIVEN: The vertices of triangle ABC are A (−2, 1) and B (5, 4) and C (2, −3)

TO FIND: The area of triangle ABC and length if the altitude through A

PROOF: We know area of triangle formed by three points is given by

Now Area of ΔABC

Taking three points A (−2, 1) and B (5, 4) and C(2, −3)

We have

Now,

#### Page No 14.48:

#### Question 5:

Show that the following sets of points are collinear.

(a) (2, 5), (4, 6) and (8, 8)

(b) (1, −1), (2, 1) and (4, 5)

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

We know area of triangle formed by three points is given by

If three points are collinear the area encompassed by them is equal to 0.

The three given points are *A*(2*,* 5)*, B*(4*,* 6) and *C*(8*,* 8). Substituting these values in the earlier mentioned formula we have,

$A=\frac{1}{2}\left|2\left(6-8\right)+4\left(8-5\right)+8\left(5-6\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|2\left(-2\right)+4\left(3\right)+8\left(-1\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-4+12-8\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-12+12\right|\phantom{\rule{0ex}{0ex}}=0$

Since the area enclosed by the three points is equal to 0, the three points need to be.

The three given points are *A*(1*,* −1)*, B*(2*,* 1) and *C*(4*,* 5). Substituting these values in the earlier mentioned formula we have,

$A=\frac{1}{2}\left|1\left(1-5\right)+2\left(5+1\right)+4\left(-1-1\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|1\left(-4\right)+2\left(6\right)+4\left(-2\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-4+12-8\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-12+12\right|\phantom{\rule{0ex}{0ex}}=0$

Since the area enclosed by the three points is equal to 0, the three points need to be.

#### Page No 14.48:

#### Question 6:

Prove that the points (a, 0), (0, b) and (1, 1) are collinear if $\frac{1}{a}+\frac{1}{b}=1$.

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

We know area of triangle formed by three points is given by

If three points are collinear the area encompassed by them is equal to 0.

The three given points are *A*(*a,*0)*, B*(0*,b*) and *C*(1*,*1).

$A=\frac{1}{2}\left|a(b-1)+1(0-b)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|ab-a-b\right|$

It is given that

So we have,

Using this in the previously arrived equation for area we have,

Since the area enclosed by the three points is equal to 0, the three points need to be.

#### Page No 14.48:

#### Question 7:

The point A divides the join of *P* (−5, 1) and *Q*(3, 5) in the ratio *k*:1. Find the two values of *k* for which the area of Δ*ABC* where *B* is (1, 5) and *C*(7, −2) is equal to 2 units.

#### Answer:

GIVEN: point A divides the line segment joining P (−5, 1) and Q (3, −5) in the ratio *k*: 1

Coordinates of point B (1, 5) and C (7, −2)

TO FIND: The value of* k*

PROOF: point A divides the line segment joining P (−5, 1) and Q (3, −5) in the ratio *k*: 1

So the coordinates of A are

We know area of triangle formed by three points is given by $\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

Now Area of ΔABC= 2 sq units.

Taking three points A , B (1, 5) and C (7, −2)

Hence

#### Page No 14.48:

#### Question 8:

The area of a triangle is 5. Two of its vertices are (2, 1) and (3, −2). The third vertex lies on y = x + 3. Find the third vertex.

#### Answer:

GIVEN: The area of triangle is 5.Two of its vertices are (2, 1) and (3, −2). The third vertex lies on *y* = *x*+3

TO FIND: The third vertex.

PROOF: Let the third vertex be (*x*, *y*)

We know area of triangle formed by three points is given by $\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

Now

Taking three points(*x*, *y*), (2, 1) and (3, −2)

Also it is given the third vertex lies on *y* = *x*+3

Substituting the value in equation (1) and (2) we get

Hence the coordinates of and

#### Page No 14.48:

#### Question 9:

If , prove that the points (*a*, a^{2}), (*b*, *b*^{2}), (*c*, *c*^{2}) can never be collinear.

#### Answer:

GIVEN: If

TO PROVE: that the points can never be collinear.

PROOF:

We know three points are collinear when

Now taking three points

Also it is given that

Hence area of triangle made by these points is never zero. Hence given points are never collinear.

#### Page No 14.48:

#### Question 10:

Four points A (6, 3), B (−3, 5), C(4, −2) and D (x, 3x) are given in such a way that $\frac{\u2206DBC}{\u2206ABC}=\frac{1}{2},$find *x*.

#### Answer:

GIVEN: four points A (6, 3), B (−3, 5) C (4, −2) and D(*x*, 3*x*) such that

TO FIND: the value of *x*

PROOF:

We know area of the triangles formed by three points is given by

Now

Area of triangle DBC taking D(*x*, 3*x*), B (−3, 5), C (4, −2)

Area of triangle ABC taking, A (6, 3), B (−3, 5), C (4, −2)

Also it is given that

Substituting the values from (1) and (2) we get

#### Page No 14.48:

#### Question 11:

For what value of *a* the point (*a*, 1), (1, −1), and (11, 4) are collinear?

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

The three given points are *A*(*a,* 1)*, B*(1*, −*1) and *C*(11*,* 4). It is also said that they are collinear and hence the area enclosed by them should be 0.

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(a\times -1+1\times 4+11\times 1\right)-\left(1\times 1+11\times -1+a\times 4\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(-a+4+11\right)-\left(1-11+4a\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(-a+15\right)-\left(-10+4a\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|-a+15+10-4a\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|-5a+25\right|\phantom{\rule{0ex}{0ex}}0=-5a+25\phantom{\rule{0ex}{0ex}}5a=25\phantom{\rule{0ex}{0ex}}a=5$

Hence the value of ‘*a*’ for which the given points are collinear is.

#### Page No 14.48:

#### Question 12:

Prove that the points (a, b), (a_{1}, b_{1}) and (a −a_{1}, b −b_{1}) are collinear if ab_{1} = a_{1}b.

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

The three given points are*,* and. If they are collinear then the area enclosed by them should be 0.

$\u2206=\frac{1}{2}\left|\left(a{b}_{1}+{a}_{1}\left(b-{b}_{1}\right)+\left(a-{a}_{1}\right)b\right)-\left({a}_{1}b+\left(a-{a}_{1}\right){b}_{1}+a\left(b-{b}_{1}\right)\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(a{b}_{1}+{a}_{1}b-{a}_{1}{b}_{1}+ab-{a}_{1}b\right)-\left({a}_{1}b+a{b}_{1}-{a}_{1}{b}_{1}+ab-a{b}_{1}\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|a{b}_{1}+{a}_{1}b-{a}_{1}{b}_{1}+ab-{a}_{1}b-{a}_{1}b-a{b}_{1}+{a}_{1}{b}_{1}-ab+a{b}_{1}\right|\phantom{\rule{0ex}{0ex}}0=a{b}_{1}-{a}_{1}b\phantom{\rule{0ex}{0ex}}a{b}_{1}={a}_{1}b$

Hence we have proved that for the given conditions to be satisfied we need to have.

#### Page No 14.48:

#### Question 13:

If three points (x_{1}, y_{1}) (x_{2}, y_{2}), (x_{3}, y_{3}) lie on the same line, prove that

$\frac{{y}_{2}-{y}_{3}}{{x}_{2}{x}_{3}}+\frac{{y}_{3}-{y}_{1}}{{x}_{3}{x}_{1}}+\frac{{y}_{1}-{y}_{2}}{{x}_{1}{x}_{2}}=0$

#### Answer:

GIVEN: If three points lie on the same line

TO PROVE:

PROOF:

We know that three points are collinear if

Hence proved.

#### Page No 14.48:

#### Question 14:

If (x, y) be on the line joining the two points (1, −3) and (−4, 2) , prove that x + y + 2= 0.

#### Answer:

Since the point (*x*, *y*) lie on the line joining the points (1, −3) and (−4, 2); the area of triangle formed by these points is 0.

That is,

Thus, the result is proved.

#### Page No 14.48:

#### Question 15:

Find the value of* k* if points A*(k*, 3), B(6, −2) and C(−3, 4) are collinear.

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

The three given points are *A*(*k,* 3)*, B*(6*, −*2) and *C*(*−*3*,* 4). It is also said that they are collinear and hence the area enclosed by them should be 0.

$\u2206=\frac{1}{2}\left|\left(k\left(-2\right)+6\times 4+\left(-3\right)\times 3\right)-\left(6\times 3+\left(-3\right)\left(-2\right)+k\times 4\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(-2k+24-9\right)-\left(18+6+4k\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|-2k+15-24-4k\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|-6k-9\right|\phantom{\rule{0ex}{0ex}}0=-6k-9\phantom{\rule{0ex}{0ex}}k=-\frac{9}{6}=-\frac{3}{2}$

Hence the value of ‘*k*’ for which the given points are collinear is.

#### Page No 14.48:

#### Question 16:

Find the value of k, if the points *A*(7, −2), *B* (5, 1) and *C *(3, 2*k*) are collinear.

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

The three given points are *A*(7*, −*2)*, B*(5*,* 1) and *C*(3*,* 2*k*). It is also said that they are collinear and hence the area enclosed by them should be 0.

$\u2206=\frac{1}{2}\left|\left(7\times 1+5\times 2k+3\times -2\right)-\left(5\times -2+3\times 1+7\times 2k\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(7+10k-6\right)-\left(-10+3+14k\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|-4k+8\right|\phantom{\rule{0ex}{0ex}}0=-4k+8\phantom{\rule{0ex}{0ex}}k=2$

Hence the value of ‘*k*’ for which the given points are collinear is.

#### Page No 14.48:

#### Question 17:

If the point P (*m*, 3) lies on the line segment joining the points $A\left(-\frac{2}{5},6\right)$and *B* (2, 8), find the value of *m*.

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

It is said that the point *P*(*m,*3) lies on the line segment joining the points* * and *B*(2*,*8). Hence we understand that these three points are collinear. So the area enclosed by them should be 0.

$\u2206=\frac{1}{2}\left|\left(-\frac{2}{5}\times 3+m\times 8+2\times 6\right)-\left(m\times 6+2\times 3+\left(-\frac{2}{5}\right)\times 8\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(-\frac{6}{5}+8m+12\right)-\left(6m+6-\frac{16}{5}\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|2m+8\right|\phantom{\rule{0ex}{0ex}}0=2m+8\phantom{\rule{0ex}{0ex}}m=-4$

Hence the value of ‘*m*’ for which the given condition is satisfied is.

#### Page No 14.48:

#### Question 18:

If R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a), then prove that *x *+ *y* = *a* + *b*.

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

It is said that the point *R*(*x, y*) lies on the line segment joining the points *P*(*a, b*)* *and *Q*(*b, a*). Hence we understand that these three points are collinear. So the area enclosed by them should be 0.

$\u2206=\frac{1}{2}\left|\left(ay+xa+{b}^{2}\right)-\left(xb+by+{a}^{2}\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|ay+xa+{b}^{2}-xb-by-{a}^{2}\right|\phantom{\rule{0ex}{0ex}}0=ay+xa+{b}^{2}-xb-by-{a}^{2}\phantom{\rule{0ex}{0ex}}{a}^{2}-{b}^{2}=ax+ay-bx-by\phantom{\rule{0ex}{0ex}}\left(a+b\right)\left(a-b\right)=\left(a-b\right)\left(x+y\right)\phantom{\rule{0ex}{0ex}}\left(a+b\right)=\left(x+y\right)\phantom{\rule{0ex}{0ex}}$

Hence under the given conditions we have proved that.

#### Page No 14.48:

#### Question 19:

Find the value of k, if the points A (8, 1) B(3, −4) and C(2, k) are collinear.

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

If three points are collinear the area encompassed by them is equal to 0.

The three given points are *A*(8*,*1)*, B*(3*,**−*4) and *C*(2*,k*). It is also said that they are collinear and hence the area enclosed by them should be 0.

$\u2206=\frac{1}{2}\left|\left(8\times -4+3\times k+2\times 1\right)-\left(3\times 1+2\times -4+8\times k\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(-32+3k+2\right)-\left(3-8+8k\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|-25-5k\right|\phantom{\rule{0ex}{0ex}}k=-5$

Hence the value of ‘*k*’ for which the given points are collinear is.

#### Page No 14.48:

#### Question 20:

Find the value of a for which the area of the triangle formed by the points A(a, 2a), B(−2, 6) and C(3, 1) is 10 square units.

#### Answer:

*A*’ encompassed by three points, and is given by the formula,

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

The three given points are *A*(*a,*2*a*)*, B*(*−*2*,*6) and *C*(3*,*1). It is also said that the area enclosed by them is 10 square units. Substituting these values in the above mentioned formula we have,

$\u2206=\frac{1}{2}\left|\left(a\times 6+\left(-2\right)\times 1+3\times 2a\right)-\left(\left(-2\right)\times 2a+3\times 6+a\times 1\right)\right|\phantom{\rule{0ex}{0ex}}10=\frac{1}{2}\left|\left(6a-2+6a\right)-\left(-4a+18+a\right)\right|\phantom{\rule{0ex}{0ex}}10=\frac{1}{2}\left|15a-20\right|\phantom{\rule{0ex}{0ex}}20=\left|15a-20\right|\phantom{\rule{0ex}{0ex}}4=\left|3a-4\right|$

We have. Hence either

Or

Hence the values of ‘*a*’ which satisfies the given conditions are.

#### Page No 14.50:

#### Question 21:

If the vertices of a triangle are (1, −3), (4, *p*) and (−9, 7) and its area is 15 sq. units, find the value(s) of *p*. [CBSE 2012]

#### Answer:

Let A(1, −3), B(4, *p*) and C(−9, 7) be the vertices of the ∆ABC.

Here, *x*_{1} = 1, *y*_{1} = −3; *x*_{2} = 4, *y*_{2} = *p *and *x*_{3} = −9, *y*_{3} = 7

ar(∆ABC) = 15 square units

$\Rightarrow \frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|=15\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\left|1\left(p-7\right)+4\left[7-\left(-3\right)\right]+\left(-9\right)\left(-3-p\right)\right|=15\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\left|p-7+40+27+9p\right|=15\phantom{\rule{0ex}{0ex}}\Rightarrow \left|10p+60\right|=30$

$\Rightarrow 10p+60=30$ or $10p+60=-30$

$\Rightarrow 10p=-30$ or $10p=-90$

$\Rightarrow p=-3$ or $p=-9$

Hence, the value of *p *is −3 or −9.

#### Page No 14.50:

#### Question 22:

Find the area of a parallelogram *ABCD* if three of its vertices are *A*(2, 4), *B*(2 + $\sqrt{3}$, 5) and *C*(2, 6). [CBSE 2013]

#### Answer:

It is given that A(2, 4), B(2 + $\sqrt{3}$, 5) and C(2, 6) are the vertices of the parallelogram ABCD.

We know that the diagonal of a parallelogram divides it into two triangles having equal area.

∴ Area of the parallogram ABCD = 2 × Area of the ∆ABC

Now,

$\mathrm{ar}\left(\u2206\mathrm{ABC}\right)=\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|2\left(5-6\right)+\left(2+\sqrt{3}\right)\left(6-4\right)+2\left(4-5\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-2+4+2\sqrt{3}-2\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 2\sqrt{3}\phantom{\rule{0ex}{0ex}}=\sqrt{3}\mathrm{square}\mathrm{units}$

∴ Area of the parallogram ABCD = 2 × Area of the ∆ABC = 2 × $\sqrt{3}$ = 2$\sqrt{3}$ square units

Hence, the area of given parallelogram is 2$\sqrt{3}$ square units.

#### Page No 14.55:

#### Question 1:

Write the distance between the points A (10 cos θ, 0) and B (0, 10 sin θ).

#### Answer:

We have to find the distance between A and B.

In general, the distance between A and B is given by,

So,

But according to the trigonometric identity,

Therefore,

#### Page No 14.55:

#### Question 2:

Write the perimeter of the triangle formed by the points *O* (0, 0), *A* (*a*, 0) and B (0, b).

#### Answer:

The distance *d* between two points and is given by the formula

The perimeter of a triangle is the sum of lengths of its sides.

The three vertices of the given triangle are O(0*, *0), A(*a, *0) and B(0*, b*).

Let us now find the lengths of the sides of the triangle.

The perimeter ‘*P*’ of the triangle is thus,

Thus the perimeter of the triangle with the given vertices is.

#### Page No 14.55:

#### Question 3:

Write the ratio in which the line segment joining points (2, 3) and (3, −2) is divided by X axis.

#### Answer:

Let P be the point of intersection of *x-*axis with the line segment joining A (2, 3) and B (3,−2) which divides the line segment AB in the ratio.

Now according to the section formula if point a point P divides a line segment joining andin the ratio m: n internally than,

Now we will use section formula as,

Now equate the y component on both the sides,

On further simplification,

So *x-*axis divides AB in the ratio

#### Page No 14.55:

#### Question 4:

What is the distance between the points (5 sin 60°, 0) and (0, 5 sin 30°)?

#### Answer:

We have to find the distance between A and B.

In general, the distance between A and B is given by,

So,

But according to the trigonometric identity,

And,

Therefore,

#### Page No 14.55:

#### Question 5:

If *A* (−1, 3) , *B*(1, −1) and *C* (5, 1) are the vertices of a triangle *ABC*, what is the length of the median through vertex *A*?

#### Answer:

We have a triangle in which the co-ordinates of the vertices are A (−1, 3) B (1,−1) and

C (5, 1). In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point D of side BC can be written as,

Now equate the individual terms to get,

So co-ordinates of D is (3, 0)

So the length of median from A to the side BC,

#### Page No 14.55:

#### Question 6:

If the distance between points (x, 0) and (0, 3) is 5, what are the values of x?

#### Answer:

We have to find the unknown *x* using the distance between A and B which is 5.In general, the distance between A and B is given by,

So,

Squaring both the sides we get,

So,

#### Page No 14.55:

#### Question 7:

What is the area of the triangle formed by the points O (0, 0), A (6, 0) and B (0, 4)?

#### Answer:

The given triangle is a right angled triangle, right angled at O. the co-ordinates of the vertices are O (0, 0) A (6, 0) and B (0, 4).

So,

Altitude is 6 units and base is 4 units.

Therefore,

#### Page No 14.55:

#### Question 8:

Write the coordinates of the point dividing line segment joining points (2, 3) and (3, 4) internally in the ratio 1 : 5.

#### Answer:

Let P be the point which divide the line segment joining A (2, 3) and B (3, 4) in the ratio 1: 5.

Now we will use section formula as,

So co-ordinate of P is

#### Page No 14.55:

#### Question 9:

If the centroid of the triangle formed by points P (a, b), Q(b, c) and R (c, a) is at the origin, what is the value of a + b + c?

#### Answer:

The co-ordinates of the vertices are (*a, b*); (*b, c*) and (*c, a*)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

Therefore,

#### Page No 14.56:

#### Question 10:

In Q. No. 9, what is the value of $\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}$?

#### Answer:

The co-ordinates of the vertices are (*a, b*); (*b, c*) and (*c, a*)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

Therefore,

We have to find the value of -

Multiply and divide it by to get,

Now as we know that if,

Then,

So,

#### Page No 14.56:

#### Question 11:

Write the coordinates of a point on X-axis which is equidistant from the points (−3, 4) and (2, 5).

#### Answer:

The distance *d* between two points and is given by the formula

Here we are to find out a point on the *x*−axis which is equidistant from both the points

A(*-*3*,*4) and B(2*,*5).

Let this point be denoted as C(*x, y*).

Since the point lies on the *x*-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘A’ and ‘B’ to ‘C’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the *x*-axis which lies at equal distances from the mentioned points is.

#### Page No 14.56:

#### Question 12:

If the mid-point of the segment joining *A* (*x*, *y* + 1) and *B* (*x* + 1, *y* + 2) is *C* $\left(\frac{3}{2},\frac{5}{2}\right)$, find *x*, *y.*

#### Answer:

It is given that mid-point of line segment joining A and B is C

In general to find the mid-point of two pointsand we use section formula as,

So,

Now equate the components separately to get,

So,

Similarly,

So,

#### Page No 14.56:

#### Question 13:

Two vertices of a triangle have coordinates (−8, 7) and (9, 4) . If the centroid of the triangle is at the origin, what are the coordinates of the third vertex?

#### Answer:

The co-ordinates of other two vertices are (−8, 7) and (9, 4)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the co-ordinate of third vertex

#### Page No 14.56:

#### Question 14:

Write the coordinates the reflections of points (3, 5) in X and Y -axes.

#### Answer:

We have to find the reflection of (3, 5) along *x-*axis and *y-*axis.

Reflection of any pointalong *x-*axis is

So reflection of (3, 5) along *x-*axis is

Similarly, reflection of any pointalong *y-*axis is

So, reflection of (3, 5) along *y-*axis is

#### Page No 14.56:

#### Question 15:

If points Q and reflections of point P (−3, 4) in X and Y axes respectively, what is QR?

#### Answer:

We have to find the reflection of (−3, 4) along *x-*axis and *y-*axis.

Reflection of any pointalong *x-*axis is

So reflection of (−3, 4) along *x-*axis is

Similarly, reflection of any pointalong *y-*axis is

So, reflection of (−3, 4) along *y-*axis is

Therefore,

#### Page No 14.56:

#### Question 16:

Write the formula for the area of the triangle having its vertices at (x_{1}, y_{1}), (x2, y_{2}) and (x_{3}, y_{3}).

#### Answer:

The formula for the area ‘A’ encompassed by three points, and is given by the formula,

The area ‘A’ encompassed by three points, and is also given by the formula,

#### Page No 14.56:

#### Question 17:

Write the condition of collinearity of points (x1, y1), (x2, y2) and (x3, y3).

#### Answer:

The condition for co linearity of three points, and is that the area enclosed by them should be equal to 0.

The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Thus for the three points to be collinear we need to have,

The area ‘A’ encompassed by three points, and is also given by the formula,

Thus for the three points to be collinear we can also have,

#### Page No 14.56:

#### Question 18:

Find the values of x for which the distance between the point P(2, −3), and Q (x, 5) is 10.

#### Answer:

It is given that distance between P (2,−3) and is 10.

In general, the distance between A and B is given by,

So,

On further simplification,

#### Page No 14.56:

#### Question 19:

Write the ratio in which the line segment doining the points A (3, −6), and B (5, 3) is divided by X-axis.

#### Answer:

Let P be the point of intersection of *x-*axis with the line segment joining A (3,−6) and B (5, 3) which divides the line segment AB in the ratio.

Now we will use section formula as,

Now equate the y component on both the sides,

On further simplification,

So *x-*axis divides AB in the ratio 2:1.

#### Page No 14.56:

#### Question 20:

Find the distance between the points $\left(-\frac{8}{5},2\right)$ and $\left(\frac{2}{5},2\right)$

#### Answer:

We have to find the distance between and .

In general, the distance between A and B is given by,

So,

#### Page No 14.56:

#### Question 21:

Find the value of *a* so that the point (3, *a*) lies on the line represented by 2*x* − 3*y** *+ 5 = 0

#### Answer:

If a point is said lie on a line represented by, then the given equation of the line should hold true when the values of the co-ordinates of the points are substituted in it.

Here it is said that the point (3*, a*) lies on the line represented by the equation.

Substituting the co-ordinates of the values in the equation of the line we have,

Thus the value of ‘*a*’ satisfying the given conditions is.

#### Page No 14.56:

#### Question 22:

What is the distance between the points *A* (*c*, 0) and *B* (0, −*c*)?

#### Answer:

We have to find the distance between and .

In general, the distance between A and B is given by,

So,

#### Page No 14.56:

#### Question 23:

If *P* (2, 6) is the mid-point of the line segment joining *A* (6, 5) and *B* (4, *y*), find *y.*

#### Answer:

It is given that mid-point of line segment joining A (6, 5) and B (4, *y*) is P (2, 6)

In general to find the mid-point of two pointsand we use section formula as,

So,

Now equate the y component to get,

So,

#### Page No 14.56:

#### Question 24:

If the distance between the points (3, 0) and (0, *y*) is 5 units and *y* is positive. then what is the value of *y*?

#### Answer:

It is given that distance between P (3, 0) and is 5.

In general, the distance between A and B is given by,

So,

On further simplification,

We will neglect the negative value. So,

#### Page No 14.56:

#### Question 25:

If *P* (*x*, 6) is the mid-point of the line segment joining *A* (6, 5) and *B* (4, *y*), find *y*.

#### Answer:

It is given that mid-point of line segment joining A (6, 5) and B (4, *y*) is

In general to find the mid-point of two pointsand we use section formula as,

So,

Now equate the y component to get,

So,

#### Page No 14.56:

#### Question 26:

If *P* (2, *p*) is the mid-point of the line segment joining the points *A* (6, −5) and *B* (−2, 11). find the value of *p*.

#### Answer:

It is given that mid-point of line segment joining A (6,−5) and B (−2, 11) is

In general to find the mid-point of two pointsand we use section formula as,

So,

Now equate the *y* component to get,

#### Page No 14.56:

#### Question 27:

If *A* (1, 2) *B* (4, 3) and *C* (6, 6) are the three vertices of a parallelogram *ABCD*, find the coordinates of fourth vertex *D*.

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (1, 2);

B (4, 3) and C (6, 6). We have to find the co-ordinates of the forth vertex.

Let the forth vertex be

Now to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

Similarly,

So the forth vertex is

#### Page No 14.57:

#### Question 1:

*Mark the correct alternative in each of the following:*

The distance between the points (cos θ, 0) and (sin θ − cos θ) is

(a) $\sqrt{3}$

(b) $\sqrt{2}$

(c) 2

(d) 1

#### Answer:

We have to find the distance between and.

In general, the distance between A and B is given by,

So,

But according to the trigonometric identity,

Therefore,

So, the answer is (b)

#### Page No 14.57:

#### Question 2:

The distance between the points (*a* cos 25°, 0) and (0, *a* cos 65°) is

(a) *a *

(b) 2*a*

(c) 3*a*

(d) None of these

#### Answer:

We have to find the distance between A(*a* cos 25°, 0) and.

In general, the distance between A and B is given by,

So,

$AB=\sqrt{{\left(0-a\mathrm{cos}25\xb0\right)}^{2}+{\left(a\mathrm{cos}65\xb0-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(a\mathrm{cos}25\xb0\right)}^{2}+{\left(a\mathrm{cos}65\xb0\right)}^{2}}$

$\mathrm{cos}25\xb0=\mathrm{sin}65\xb0\mathrm{and}\mathrm{cos}65\xb0=\mathrm{sin}25\xb0$

But according to the trigonometric identity,

Therefore,

So, the answer is (a)

#### Page No 14.57:

#### Question 3:

If x is a positive integer such that the distance between points P (x, 2) and Q (3, −6) is 10 units, then *x* =

(a) 3

(b) −3

(c) 9

(d) −9

#### Answer:

It is given that distance between P (*x*, 2) and is 10.

In general, the distance between A and B is given by,

So,

On further simplification,

We will neglect the negative value. So,

So the answer is (c)

#### Page No 14.57:

#### Question 4:

The distance between the points (a cos θ + b sin θ, 0) and (0, a sin θ − b cos θ) is

(a) a^{2} + b^{2}

(b) a + b

(c) a^{2} − b^{2}

(d) $\sqrt{a2+b2}$

#### Answer:

We have to find the distance between and.

In general, the distance between A and B is given by,

So,

But according to the trigonometric identity,

Therefore,

So, the answer is (d)

#### Page No 14.57:

#### Question 5:

If the distance between the points (4, *p*) and (1, 0) is 5, then *p = *

(a) ± 4

(b) 4

(c) −4

(d) 0* *

#### Answer:

It is given that distance between P (4, *p*) and is 5.

In general, the distance between A and B is given by,

So,

On further simplification,

So,

So the answer is (a)

#### Page No 14.57:

#### Question 6:

A line segment is of length 10 units. If the coordinates of its one end are (2, −3) and the abscissa of the other end is 10, then its ordinate is

(a) 9, 6

(b) 3, −9

(c) −3, 9

(d) 9, −6

#### Answer:

It is given that distance between P (2,−3) and is 10.

In general, the distance between A and B is given by,

So,

On further simplification,

We will neglect the negative value. So,

So the answer is (b)

#### Page No 14.57:

#### Question 7:

The perimeter of the triangle formed by the points (0, 0), (0, 1) and (0, 1) is

(a) 1 ± $\sqrt{2}$

(b) $\sqrt{2}$ + 1

(c) 3

(d) $2+\sqrt{2}$

#### Answer:

We have a triangle whose co-ordinates are A (0, 0); B (1, 0); C (0, 1). So clearly the triangle is right angled triangle, right angled at A. So,

Now apply Pythagoras theorem to get the hypotenuse,

So the perimeter of the triangle is,

Therefore the answer is (d)

#### Page No 14.57:

#### Question 8:

If *A* (2, 2), *B* (−4, −4) and *C* (5, −8) are the vertices of a triangle, than the length of the median through vertex *C* is

(a) $\sqrt{65}$

(b) $\sqrt{117}$

(c) $\sqrt{85}$

(d) $\sqrt{113}$

#### Answer:

We have a triangle in which the co-ordinates of the vertices are A (2, 2) B (−4,−4) and C (5,−8).

In general to find the mid-point of two points and we use section formula as,

Therefore mid-point D of side AB can be written as,

Now equate the individual terms to get,

So co-ordinates of D is (−1,−1)

So the length of median from C to the side AB,

So the answer is (c)

#### Page No 14.57:

#### Question 9:

If three points (0, 0), $\left(3,\sqrt{3}\right)$ and (3, λ) form an equilateral triangle, then λ =

(a) 2

(b) −3

(c) −4

(d) None of these

#### Answer:

We have an equilateral triangle whose co-ordinates are A (0, 0); and.

Since the triangle is equilateral. So,

So,

Cancel out the common terms from both the sides,

Therefore,

So, the answer is (d)

#### Page No 14.57:

#### Question 10:

If the points (*k*, 2*k*), (3*k*, 3*k*) and (3, 1) are collinear, then *k*

(a) $\frac{1}{3}$

(b) $-\frac{1}{3}$

(c) $\frac{2}{3}$

(d) $-\frac{2}{3}$

#### Answer:

We have three collinear points.

In general if are collinear then, area of the triangle is 0.

So,

So,

Take out the common terms,

Therefore,

So the answer is (b)

#### Page No 14.57:

#### Question 11:

The coordinates of the point on X-axis which are equidistant from the points (−3, 4) and (2, 5) are

(a) (20, 0)

(b) (−23, 0)

(c) $\left(\frac{4}{5},0\right)$

(d) None of these

#### Answer:

Let the point be A be equidistant from the two given points P (−3, 4) and Q (2, 5).

So applying distance formula, we get,

Therefore,

Hence the co-ordinates of A are

So the answer is- (D) none of these.

#### Page No 14.57:

#### Question 12:

If (−1, 2), (2, −1) and (3, 1) are any three vertices of a parallelogram, then

(a) *a* = 2, *b* = 0

(b)* a*= −2, *b* = 0

(c) *a* = −2, *b *= 6

(d) *a* = 6, *b* = 2

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (−1, 2);

B (2,−1) and C(3, 1). We have to find the co-ordinates of the forth vertex.

Let the forth vertex be

Now to find the mid-point of two points and we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

Similarly,

So the forth vertex is

#### Page No 14.58:

#### Question 13:

If *A* (5, 3), *B* (11, −5) and *P* (12, *y*) are the vertices of a right triangle right angled at *P*, then *y*=

(a) −2, 4

(b) −2, −4

(c) 2, −4

(d) 2, 4

#### Answer:

Disclaimer: option (b) and (c) are given to be same in the book. So, we are considering option (b) as −2, −4

instead of −2, 4.

We have a right angled triangle whose co-ordinates are A (5, 3); B (11,−5);

. So clearly the triangle is, right angled at A. So,

Now apply Pythagoras theorem to get,

So,

On further simplification we get the quadratic equation as,

Now solve this equation using factorization method to get,

Therefore,

So the answer is (c)

#### Page No 14.58:

#### Question 14:

The area of the triangle formed by (*a*, *b* + *c*), (*b*, *c* + *a*) and (*c*, *a* + *b*)

(a) *a* + *b* + *c*

(b) *abc*

(c) (*a* +* b* + *c*)^{2}

(d) 0

#### Answer:

We have three non-collinear points.

In general if are non-collinear points then are of the triangle formed is given b*y-*

So,

So the answer is (d)

#### Page No 14.58:

#### Question 15:

If (*x* , 2), (−3, −4) and (7, −5) are collinear, then* x* =

(a) 60

(b) 63

(c) −63

(d) −60

#### Answer:

We have three collinear points.

In general if are collinear then,

So,

So,

Therefore,

So the answer is (c)

#### Page No 14.58:

#### Question 16:

If points (*t*, 2*t*), (−2, 6) and (3, 1) are collinear, then *t* =

(a) $\frac{3}{4}$

(b) $\frac{4}{3}$

(c) $\frac{5}{3}$

(d) $\frac{3}{5}$

#### Answer:

We have three collinear points.

In general if are collinear then,

So,

So,

So,

Therefore,

So the answer is (b)

#### Page No 14.58:

#### Question 17:

If the area of the triangle formed by the points (*x*, 2*x*), (−2, 6) and (3, 1) is 5 square units , then x =

(a) $\frac{2}{3}$

(b) $\frac{3}{5}$

(c) 3

(d) 5

#### Answer:

We have the co-ordinates of the vertices of the triangle aswhich has an area of 5 sq.units.

In general if are non-collinear points then area of the triangle formed is given b*y-*,

So,

Simplify the modulus function to get,

Therefore,

So the answer is (a)

#### Page No 14.58:

#### Question 18:

If points *(a*, 0), (0, *b*) and (1, 1) are collinear, then $\frac{1}{a}+\frac{1}{b}=$

(a) 1

(b) 2

(c) 0

(d) −1

#### Answer:

We have three collinear points.

In general if are collinear then,

So,

So,

Divide both the sides by,

So the answer is (a)

#### Page No 14.58:

#### Question 19:

If the centroid of a triangle is (1, 4) and two of its vertices are (4, −3) and (−9, 7), then the area of the triangle is

(a) 183 sq. units

(b) $\frac{183}{2}$ sq. units

(c) 366 sq. units

(d) $\frac{183}{4}$ sq. units

#### Answer:

The co-ordinates of other two vertices are (4,−3) and (−9, 7)

The co-ordinate of the centroid is (1, 4)

We know that the co-ordinates of the centroid of a triangle whose vertices are is

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the co-ordinate of third vertex is (8, 8)

In general if are non-collinear points then are of the triangle formed is given b*y-*,

So,

So the answer is (b)

#### Page No 14.58:

#### Question 20:

The line segment joining points (−3, −4), and (1, −2) is divided by y-axis in the ratio

(a) 1 : 3

(b) 2 : 3

(c) 3 : 1

(d) 2 : 3

#### Answer:

Let P be the point of intersection of *y-*axis with the line segment joining A (−3,−4) and B (1,−2) which divides the line segment AB in the ratio.

Now according to the section formula if point a point P divides a line segment joining and in the ratio *m:n* internally than,

Now we will use section formula as,

Now equate the *x* component on both the sides,

On further simplification,

So *y-*axis divides AB in the ratio

So the answer is (c)

#### Page No 14.58:

#### Question 21:

The ratio in which (4, 5) divides the join of (2, 3) and (7, 8) is

(a) −2 : 3

(b) −3 : 2

(c) 3 : 2

(d) 2 : 3

#### Answer:

Here it is said that the point (4, 5)* *divides the points A(2,3) and B(7,8). Substituting these values in the above formula we have,

Equating the individual components we have,

Hence the correct choice is option (d).

#### Page No 14.58:

#### Question 22:

The ratio in which the x-axis divides the segment joining (3, 6) and (12, −3) is

(a) 2: 1

(b) 1 : 2

(c) −2 : 1

(d) 1 : −2

#### Answer:

Let P be the point of intersection of *x-*axis with the line segment joining A (3, 6) and B (12, −3) which divides the line segment AB in the ratio.

Now we will use section formula as,

Now equate the y component on both the sides,

On further simplification,

So *x-*axis divides AB in the ratio

So the answer is (a)

#### Page No 14.58:

#### Question 23:

If the centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is at the origin, then *a*^{3}^{ }+ *b*^{3} + *c*^{3} =

(a) *abc*

(b) 0

(c) *a* + *b* + *c*

(d) 3 *abc*

#### Answer:

The co-ordinates of the vertices are (*a, b*); (*b, c*) and (*c, a*)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are is

So,

Compare individual terms on both the sides-

Therefore,

We have to find the value of -

Now as we know that if,

Then,

So the answer is (d)

#### Page No 14.58:

#### Question 24:

If Points (1, 2) (−5, 6) and (a, −2) are collinear, then a =

(a) −3

(b) 7

(c) 2

(d) −2

#### Answer:

We have three collinear points.

In general if are collinear then,

So,

So,

Therefore,

So the answer is (b)

#### Page No 14.58:

#### Question 25:

If the centroid of the triangle formed by (7, x) (y, −6) and (9, 10) is at (6, 3), then (x, y) =

(a) (4, 5)

(b) (5, 4)

(c) (−5, −2)

(d) (5, 2)

#### Answer:

We have to find the unknown co-ordinates.

The co-ordinates of vertices are

The co-ordinate of the centroid is (6, 3)

We know that the co-ordinates of the centroid of a triangle whose vertices are is

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the answer is (d)

#### Page No 14.58:

#### Question 26:

The distance of the point (4, 7) from the x-axis is

(a) 4

(b) 7

(c) 11

(d) $\sqrt{65}$

#### Answer:

The ordinate of a point gives its distance from the *x-*axis.

So, the distance of (4, 7) from *x-*axis is

So the answer is (b)

#### Page No 14.58:

#### Question 27:

The distance of the point (4, 7) from the y-axis is

(a) 4

(b) 7

(c) 11

(d) $\sqrt{65}$

#### Answer:

The distance of a point from *y-*axis is given by abscissa of that point.

So, distance of (4, 7) from *y-*axis is.

So the answer is (a)

#### Page No 14.58:

#### Question 28:

If *P* is a point on x-axis such that its distance from the origin is 3 units, then the coordinates of a point *Q *on *OY* such that *OP* = *OQ*, are

(a) (0, 3)

(b) (3, 0)

(c) (0, 0)

(d) (0, −3)

#### Answer:

GIVEN: If P is a point on *x* axis such that its distance from the origin is 3 units.

TO FIND: The coordinates of a point Q on OY such that OP= OQ.

On *x* axis y coordinates is 0. Hence the coordinates of point P will be (3, 0) as it is given that the distance from origin is 3 units.

Now then the coordinates of Q on OY such that OP = OQ

On *y* axis *x* coordinates is 0. Hence the coordinates of point Q will be (0, 3)

Hence correct option is (*a*)

#### Page No 14.59:

#### Question 29:

If the points(*x*, 4) lies on a circle whose centre is at the origin and radius is 5, then *x* =

(a) ±5

(b) ±3

(c) 0

(d) ±4

#### Answer:

It is given that the point A(*x*, 4) is at a distance of 5 units from origin O.

So, apply the distance formula to get,

Therefore,

So,

So the answer is (b)

#### Page No 14.59:

#### Question 30:

If the points *P* (*x*, *y*) is equidistant from *A* (5, 1) and *B* (−1*,* 5*)*, then

(a) 5*x* = *y*

(b) *x* = 5*y*

(c) 3*x* = 2*y*

(d) 2*x* = 3*y*

#### Answer:

It is given that is equidistant to the point

So,

So apply distance formula to get the co-ordinates of the unknown value as,

On further simplification we get,

So,

Thus,

So the answer is (c)

#### Page No 14.59:

#### Question 31:

If points *A* (5, *p*) *B* (1, 5), *C* (2, 1) and *D* (6, 2) form a square *ABCD*, then *p* =

(a) 7

(b) 3

(c) 6

(d) 8

#### Answer:

The distance *d* between two points and is given by the formula

In a square all the sides are equal to each other.

Here the four points are A(5,*p*)*, *B(1,5), C(2,1) and D(6,2).

The vertex ‘A’ should be equidistant from ‘B’ as well as *‘*D’

Let us now find out the distances ‘AB’ and ‘AD’.

These two need to be equal.

Equating the above two equations we have,

Squaring on both sides we have,

Hence the correct choice is option (c).

#### Page No 14.59:

#### Question 32:

The coordinates of the circumcentre of the triangle formed by the points *O* (0, 0), *A* (a, 0 and *B* (0, *b*) are

(a) (*a*, *b*)

(b) $\left(\frac{a}{2},\frac{b}{2}\right)$

(c) $\left(\frac{b}{2},\frac{a}{2}\right)$

(d) (*b*, *a*)

#### Answer:

The distance *d* between two points and is given by the formula

The circumcentre of a triangle is the point which is equidistant from each of the three vertices of the triangle.

Here the three vertices of the triangle are given to be O(0,0), A(*a*,0) and B(0,*b*).

Let the circumcentre of the triangle be represented by the point R*(x, y)*.

So we have

Equating the first pair of these equations we have,

Squaring on both sides of the equation we have,

Equating another pair of the equations we have,

Squaring on both sides of the equation we have,

Hence the correct choice is option (*b*).

#### Page No 14.59:

#### Question 33:

The coordinates of a point on x-axis which lies on the perpendicular bisector of the line segment joining the points (7, 6) and (−3, 4) are

(a) (0, 2)

(b) (3, 0)

(c) (0, 3)

(d) (2, 0)

#### Answer:

TO FIND: The coordinates of a point on *x* axis which lies on perpendicular bisector of line segment joining points (7, 6) and (−3, 4).

Let P(*x*, *y*) be any point on the perpendicular bisector of AB. Then,

PA=PB

On *x*-axis *y* is 0, so substituting *y*=0 we get *x*= 3

Hence the coordinates of point is i.e. option (*b*) is correct

#### Page No 14.59:

#### Question 34:

If the centroid of the triangle formed by the points (3, −5), (−7, 4), (10, −*k*) is at the point (*k* −1), then *k* =

(a) 3

(b) 1

(c) 2

(d) 4

#### Answer:

We have to find the unknown co-ordinates.

The co-ordinates of vertices are

The co-ordinate of the centroid is

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

So the answer is (c)

#### Page No 14.59:

#### Question 35:

If (−2, 1) is the centroid of the triangle having its vertices at (*x* , 0) (5, −2), (−8, *y*), then *x*, *y* satisfy the relation

(a) 3*x* + 8*y* = 0

(b) 3*x* − 8*y* = 0

(c) 8*x* + 3*y* = 0

(d) 8*x* = 3*y*

#### Answer:

We have to find the unknown co-ordinates.

The co-ordinates of vertices are

The co-ordinate of the centroid is (−2, 1)

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

It can be observed that (*x*, *y*) = (−3, 5) does not satisfy any of the relations 3*x* + 8*y* = 0, 3*x* − 8*y* = 0, 8*x* + 3*y* = 0 or 8*x* = 3*y*.

#### Page No 14.59:

#### Question 36:

The coordinates of the fourth vertex of the rectangle formed by the points (0, 0), (2, 0), (0, 3) are

(a) (3, 0)

(b) (0, 2)

(c) (2, 3)

(d) (3, 2)

#### Answer:

We have to find the co-ordinates of forth vertex of the rectangle ABCD.

We the co-ordinates of the vertices as (0, 0); (2, 0); (0, 3)

Rectangle has opposite pair of sides equal.

When we plot the given co-ordinates of the vertices on a Cartesian plane, we observe that the length and width of the rectangle is 2 and 3 units respectively.

So the co-ordinate of the forth vertex is

So the answer is (c).

#### Page No 14.59:

#### Question 37:

The length of a line segment joining *A* (2, −3) and *B* is 10 units. If the abscissa of *B* is 10 units, then its ordinates can be

(a) 3 or −9

(b) −3 or 9

(c) 6 or 27

(d) −6 or −27

#### Answer:

It is given that distance between P (2,−3) and is 10.

In general, the distance between A and B is given by,

So,

On further simplification,

We will neglect the negative value. So,

So the answer is (a)

#### Page No 14.59:

#### Question 38:

The ratio in which the line segment joining *P* (*x*_{1}, *y*_{1}) and *Q* (*x*_{2},* **y*_{2}) is divided by x-axis is

(a) y_{1} : y_{2}

(b) −y_{1} : y_{2}

(c) x_{1} : x_{2}

(d) −x_{1} : x_{2}

#### Answer:

Let C be the point of intersection of *x-*axis with the line segment joining and which divides the line segment PQ in the ratio.

Now according to the section formula if point a point P divides a line segment joining andin the ratio *m:n* internally than,

Now we will use section formula as,

Now equate the y component on both the sides,

On further simplification,

So the answer is (b)

#### Page No 14.59:

#### Question 39:

The ratio in which the line segment joining points *A* (*a*_{1}, *b*_{1}) and *B* (*a*_{2}, *b*_{2}) is divided by *y*-axis is

(a) −*a*_{1} : *a*_{2}

(b) *a*_{1}_{ }: *a*_{2}

(c) *b*_{1} : *b*_{2}

(d) −*b*_{1} : *b*_{2}

#### Answer:

Let P be the point of intersection of *y-*axis with the line segment joiningandwhich divides the line segment AB in the ratio.

Now according to the section formula if point a point P divides a line segment joining andin the ratio *m:n* internally than,

Now we will use section formula as,

Now equate the *x* component on both the sides,

On further simplification,

So the answer is (a)

#### Page No 14.59:

#### Question 40:

If the line segment joining the points (3, −4), and (1, 2) is trisected at points *P* (a, −2) and *Q* $\left(\frac{5}{3},b\right)$, Then,

(a) $a=\frac{8}{3},b=\frac{2}{3}$

(b) $a=\frac{7}{3},b=0$

(c) $a=\frac{1}{3},b=1$

(d) $a=\frac{2}{3},b=\frac{1}{3}$

#### Answer:

We have two points A (3,−4) and B (1, 2). There are two points P (*a*,−2) and Q which trisect the line segment joining A and B.

The point P is the point of trisection of the line segment AB. So, P divides AB in the ratio 1: 2

Now we will use section formula to find the co-ordinates of unknown point A as,

Equate the individual terms on both the sides. We get,

Similarly, the point Q is the point of trisection of the line segment AB. So, Q divides AB in the ratio 2: 1

Now we will use section formula to find the co-ordinates of unknown point A as,

Equate the individual terms on both the sides. We get,

So the answer is (b)

#### Page No 14.61:

#### Question 41:

If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (−2, 5), then the coordinates of the other end of the diameter are

(a) (−6, 7) (b) (6, −7) (c) (6, 7) (d) (−6,−7) [CBSE 2012]

#### Answer:

Let O(−2, 5) be the centre of the given circle and A(2, 3) and B(*x*, *y*) be the end points of a diameter of the circle.

Then, O is the mid-point of AB.

Using mid-point formula, we have

$\therefore \frac{2+x}{2}=-2$ and $\frac{3+y}{2}=5$

$\Rightarrow 2+x=-4$ and $3+y=10$

$\Rightarrow x=-6$ and $y=7$

Thus, the coordinates of the other end of the diameter are (−6, 7).

Hence, the correct answer is option A.

#### Page No 14.61:

#### Question 42:

The coordinates of the point P dividing the line segment joining the points *A* (1, 3) and *B* (4, 6) in the ratio 2 : 1 are

(a) (2, 4) (b) (3, 5) (c) (4, 2) (d) (5, 3) [CBSE 2012]

#### Answer:

It is given that P divides the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1.

Using section formula, we get

Coordinates of P$=\left(\frac{2\times 4+1\times 1}{2+1},\frac{2\times 6+1\times 3}{2+1}\right)=\left(\frac{9}{3},\frac{15}{3}\right)=\left(3,5\right)$

Thus, the coordinates of P are (3, 5).

Hence, the correct answer is option B.

#### Page No 14.61:

#### Question 43:

In Fig. 14.46, the area of Δ*ABC* (in square units) is

(a) 15 (b) 10 (c) 7.5 (d) 2.5 [CBSE 2013]

#### Answer:

The coordinates of A are (1, 3).

∴ Distance of A from the *x*-axis, AD = *y*-coordinate of A = 3 units

The number of units between B and C on the *x*-axis are 5.

∴ BC = 5 units

Now,

Area of ∆ABC = $\frac{1}{2}\times \mathrm{BC}\times \mathrm{AD}=\frac{1}{2}\times 5\times 3=\frac{15}{2}=7.5$ square units

Thus, the area of ∆ABC is 7.5 square units.

Hence, the correct answer is option C.

#### Page No 14.61:

#### Question 44:

The point on the *x*-axis which is equidistant from points (−1, 0) and (5, 0) is

(a) (0, 2) (b) (2, 0) (c) (3, 0) (d) (0, 3) [CBSE 2013]

#### Answer:

Let A(−1, 0) and B(5, 0) be the given points. Suppose the required point on the *x*-axis be P(*x*, 0).

It is given that P(*x*, 0) is equidistant from A(−1, 0) and B(5, 0).

∴ PA = PB

⇒ PA^{2 }= PB^{2}

$\Rightarrow {\left[x-\left(-1\right)\right]}^{2}+{\left(0-0\right)}^{2}={\left(x-5\right)}^{2}+{\left(0-0\right)}^{2}$ (Using distance formula)

$\Rightarrow {\left(x+1\right)}^{2}={\left(x-5\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+2x+1={x}^{2}-10x+25\phantom{\rule{0ex}{0ex}}\Rightarrow 12x=24\phantom{\rule{0ex}{0ex}}\Rightarrow x=2$

Thus, the required point is (2, 0).

Hence, the correct answer is option B.

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