Mathematics Part ii Solutions Solutions for Class 10 Math Chapter 4 Geometry And Algebra are provided here with simple step-by-step explanations. These solutions for Geometry And Algebra are extremely popular among class 10 students for Math Geometry And Algebra Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part ii Solutions Book of class 10 Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Mathematics Part ii Solutions Solutions. All Mathematics Part ii Solutions Solutions for class 10 Math are prepared by experts and are 100% accurate.

#### Page No 181:

#### Question 1:

The centre of a circle is (3, 4) and it passes through the point (2, 5). What is its radius?

#### Answer:

Let the centre of the circle be denoted by point *O* and the point through which the circle passes be denoted by *A*.

Given: Coordinates of point *O* and *A* are (3, 4) and (2, 5) respectively.

Radius of the circle = Distance between the points *O* and *A*

We know that the distance between two points is given by:

D =

Now, *OA* =

Thus, the radius of the circle is units.

#### Page No 181:

#### Question 2:

A circle of radius 3 is drawn with centre at (−2, 1). Find out whether the point (4, 1) lies on the circle, within the circle or outside the circle.

#### Answer:

Let the centre of the circle be denoted by point *O* and the given point be denoted by *A*.

Given: Coordinates of point *O* and *A* are (−2, 1) and (4, 1) respectively.

Radius of the circle = 3 units

We know that distance between two points is given by:

D =

Now, *OA* =

So, the distance between the points *O* and *A* is 6 units, but the radius of the circle is 3 units.

Thus, the point (4, 1) lie outside the circle with centre at (−2, 1).

#### Page No 181:

#### Question 3:

Prove that we get a right angled triangle by joining the points (2, 1), (3, 4), (−3, 6).

#### Answer:

Let the given points be *A* (2, 1), *B* (3, 4) and *C* (−3, 6).

We know that distance between two points is given by:

D =

Now, *AB *=

⇒ *AB*^{2} = 10 ……… (1)

*BC* =

⇒ *BC*^{2} = 40 ……… (2)

*CA* =

*CA*^{2} = 50 ……… (3)

From (1), (2) and (3), we have:

50 = 40 + 10

i.e. *CA*^{2}^{ }= *BC*^{2} + *AB*^{2}

Therefore, the points *A*, *B *and *C* follow the Pythagoras theorem.

Thus, the points *A*, *B* and *C* form a right angled triangle, right-angled at *B*.

#### Page No 181:

#### Question 4:

How many points are there on the *x*-axis, at a distance 5 from the point (1, 3)? What are their coordinates? What about such points on the *y*-axis?

#### Answer:

Let the given point be* A* (1, 3).

**On ****x****-axis:**

Let the coordinates of the required points be (*h*, *k*). But since the points lie on the *x*-axis, the *y*-coordinate should be zero.

So, the coordinates of the required points are (*h*, 0).

We know that distance between two points is given by:

D =

The required point is at a distance of 5 units from the point *A*

Thus, there are two points on the *x *-axis which are at a distance of 5 units from the point (1, 3) and their coordinates are (−3, 0) and (5, 0).

**On ****y****-axis****:**

Let the coordinates of the required points be (*h*, *k*). But since the points lie on the *y*-axis, the *x*-coordinate should be zero.

So, the coordinates of the required points are (0, *k*).

The required point is at a distance of 5 units from the point *A*

Thus, there are two points on *y*-axis which are at a distance of 5 units from the point (1, 3) and their coordinates are

#### Page No 182:

#### Question 1:

Find the coordinates of the points B, C, D in the picture below:

Write the lengths AB, BC, CD in the order of their magnitudes.

#### Answer:

For knowing the coordinates of points *B*, *C* and *D*, we need to know the length of the line segments *BA*, *CA* and *DA* respectively.

We know that distance between two points is given by:

D =

Now, *OA *=

Δ*BAO* is a right angled triangle, right angled at *A*.

Here, ∠*BOA* = 30°

∴ Tan 30° =

Also, Δ*CAO* is a right angled triangle, right angled at *A*.

Here, ∠*COA* = 30° + 15° = 45°

∴ Tan 45° =

Δ*DAO* is a right angled triangle, right angled at* A*.

Here, ∠*DOA* = 30° + 15° + 15° = 60°

∴ Tan 60° =

Points *A*, *B*, *C* and *D* all lie on a line parallel to *y*-axis.

Since, the distance between point *A* and *y*-axis is 9 units, the distance between points *B*, *C* and *D* and *y*-axis is also 9 units.

Thus, the coordinates of points *B*, *C *and *D* are (9, ), (9, 9) and (9, ) respectively.

Now, *AB* =

*BC* = *AC* − *AB* =

*CD* = *AD* − *AC *=

Thus, the lengths of *AB*, *BC* and *CD* in order of their magnitudes are:

#### Page No 182:

#### Question 2:

The circle centred at (2, 3) and of radius 5, intersects the *x*-axis at A and B. Find the coordinates of A and B and also the length of the chord AB.

#### Answer:

Coordinates of the centre of the circle = (2, 3)

Radius of the circle = 5 units

Let the points of intersection of the circle on the *x*- axis be *A* (*h*, 0) and *B* (*k*, 0).

We know that distance between two points is given by:

D =

Also, the distance of the required points from the centre of the circle is 5 units.

Therefore, the coordinates of the required points are *A* (6, 0) and *B* (−2, 0).

Distance between the points AB

_{Thus, the length of chord AB is 8 units.}

#### Page No 182:

#### Question 3:

The vertices of a triangle are the points (1, 2), (2, 3), (3, 1). Find the centre and radius of its circumcircle.

#### Answer:

Let the coordinates of the triangle be *A* (1, 2), *B *(2, 3) and *C *(3, 1).

If a circle is circumscribing the triangle *ABC,* then it should pass through the points *A*, *B* and *C* and the centre of this circle, say *O* should be equidistant from the points *A*,* B *and *C*.

⇒ *OA* = *OB* = *OC*

Let the coordinates of point *O* be (*h*, *k*).

We know that distance between two points is given by:

D =

⇒ *OA* =

⇒ *OB *=

⇒ *OC *=

*OA* = *OB*

⇒=

Also, *OB* = *OC*

⇒=

Solving (1) and (2), we get:

Thus, the coordinates of the centre of the circumcircle are

Radius of the circle = *OA*

_{Thus, the radius of the circumcircle is }

#### Page No 187:

#### Question 1:

Does the line joining (2, 3) and (3, 1) pass through the point (5, 6)? What about (5, −9)?

#### Answer:

Let the given points be *A* (2, 3), *B* (3, −1) and *C* (5, 6).

For a line joining the points* A *and *B* to pass through the point *C*, the slope of the line *BC* must be equal to the slope of line *AC*.

Slope of line *BC* =

Slope of line *AC *=

⇒ Slope of line *BC* ≠ Slope of line *AC*

Therefore, the line joining the points (2, 3) and (3, −1) does not pass through the point (5, 6).

Now, consider the point *D* (5, −9).

For a line joining the points *A* and *B* to pass through the point *D*, the slope of the line *BD* must be equal to slope of line *AD*.

Slope of line *BD* =

Slope of line *AD* =

⇒ Slope of line *BD* = Slope of line *AD*

Therefore, the line joining the points (2, 3) and (3, −1) passes through the point (5, −9).

#### Page No 187:

#### Question 2:

Prove that the points (1, 4), (4, 1) and lie on the same line.

#### Answer:

Let the given points be *A* (1, 4), *B* (4, 1) and *C *

For the points *A*, *B* and *C* to lie on the same line, the line joining the points *A* and *B* should pass through the point *C*. So, the slope of the line *BC* should be equal to the slope of the line *AC*.

Slope of line* BC* =

Slope of line *AC* =

⇒ Slope of line *BC* = Slope of line *AC*

Therefore, the points (1, 4), (4, 1) and lie on the same line.

#### Page No 187:

#### Question 3:

Prove that the points (2, 3), (7, 5), (9, 8), (4, 6) are the vertices of a parallelogram.

#### Answer:

Let the points be *A* (2, 3), *B* (7, 5), *C* (9, 8) and *D* (4, 6).

For the given four points to form a parallelogram, the opposite sides must be parallel, i.e., * AB* should be parallel to *DC* and* BC* should be parallel to * AD*.

Slope of line *AB *=

Slope of line *DC *=

Slope of line *BC *=

Slope of line *AD *=

⇒ Slope of line *AB *= Slope of line *DC*

Slope of line *BC *= Slope of line *AD*

Thus, the given vertices form a parallelogram.

#### Page No 187:

#### Question 4:

Prove that the line joining the points (2, 1) and (1, 2) and the line joining the points (3, 5) and (4, 7) are not parallel. What are the coordinates of their point of intersection?

#### Answer:

Let the given points be *A* (2, 1), *B* (1, 2), *C *(3, 5) and *D* (4, 7).

Slope of line *AB *=

Slope of line* CD* =

⇒ Slope of line *AB* ≠ Slope of line *CD*

Therefore, the given lines are not parallel.

Let the point of intersection of both the lines be *E *(*x*_{1}, *y*_{1}).

Therefore, the slope of the line* BE *should be equal to the slope of the line *AB*.

Slope of line *BE *=

⇒ *y*_{1}_{ }− 2 = −*x*_{1} + 1

⇒ *y*_{1}_{ }+ *x*_{1} = 3 ….. (1)

Similarly, the slope of the line *DE* should be equal to the slope of the line *CD.*

Slope of line *DE* =

⇒ *y*_{1}−7 = 2*x*_{1} − 8

⇒ *y*_{1}− 2*x*_{1} = −1 ….. (2)

Solving (1) and (2), we get:

*x*_{1} = , _{ }*y*_{1} =

Therefore, the point of intersection of both the lines is

#### Page No 187:

#### Question 5:

Write down the coordinates of two more points on the line through (1, 3) of slope.

#### Answer:

Let the given point be *A* (1, 3).

Let the point *B* (*x*, *y*) lie on the line passing through point *A*.

Given that slope of the line passing through point *A *is

⇒

⇒ 2*y* − 6 =* x* − 1

⇒ *x* − 2*y*_{ }+ 5 =* *0

This is the equation of the line passing through point *A* and having slope

Any point on this line can be found out by assuming *x* to have any value, say 2 or 3 and putting it in the above equation.

2 − 2*y*_{ }+ 5 =* *0

⇒ −2*y*_{ }+ 7 = 0

⇒ *y* =

Also, 3 − 2*y*_{ }+ 5 =* *0

⇒ −2*y*_{ }+ 8 = 0

⇒ *y* = 4

Thus, the coordinates of two more points on this line are and (3, 4).

#### Page No 187:

#### Question 6:

Two lines are drawn through the point (1, 3) one of slope and the other of slope −2. Write the coordinates of one more point on each of these lines. Prove that these lines are perpendicular to each other.

#### Answer:

Let the given point be *A* (1, 3).

Let the point *B* (*x*, *y*) lie on the line passing through point *A*.

Given that the slope of the first line passing through point *A *is

⇒

⇒ 2*y* − 6 =* x* − 1

⇒ *x* − 2*y*_{ }+ 5 =* *0

This is the equation of the line passing through point *A* and having slope

Any point on this line can be found out by assuming *x* to have any value, say 2 and putting it in the above equation.

2 − 2*y*_{ }+ 5 =* *0

⇒ −2*y*_{ }+ 7 = 0

⇒ *y* =

Thus, the coordinates of one more point on the first line are

Also, given that the slope of the second line passing through point *A *is −2.

⇒

⇒ *y* − 3 =* *−2*x* + 2

⇒ 2*x* + *y*_{ }− 5 =* *0

This is the equation of the line passing through point *A* and having slope −2.

Any point on this line can be found out by assuming *x* to have any value, say 2 and putting it in the above equation.

4 + *y*_{ }− 5 =* *0

⇒ *y*_{ }− 1 = 0

⇒ *y* = 1

Thus, the coordinates of one more point on the second line are (2, 1).

Let *m*_{1} = , *m*_{2} = −2

For two lines to be perpendicular, the product of their slopes should be equal to −1.

*m*_{1} *m*_{2} = × (−2) = −1

Therefore, the given lines are perpendicular to each other.

#### Page No 190:

#### Question 1:

Prove that for all points on the line joining the origin and the point (4, 2) the *x*-coordinates is double the *y*-coordinates. What is the equation of this line?

#### Answer:

The equation of a line passing through points (*x*_{1}, *y*_{1}) and (*x*_{2}, *y*_{2}) is given by:

The equation of the line passing through the points (0, 0) and (4, 2) is:

⇒ 4*y* = 2*x*

⇒ 2*y* = *x*

This is the equation of the line passing through the points (0, 0) and (4, 2).

Observing the equation, we can say that the *x*-coordinate of all the points on the line joining the given points is double the *y*-coordinate.

#### Page No 190:

#### Question 2:

What is the equation of the line joining the points (1, 3) and (2, 7)? Prove that if (*x*, *y*) is a point on this line, so is the point (*x* + 1, *y* + 4).

#### Answer:

The equation of a line passing through points (*x*_{1}, *y*_{1}) and (*x*_{2}, *y*_{2}) is given by:

The equation of the line passing through points (1, 3) and (2, 7) is:

⇒ *y* − 3 = 4*x* − 4

⇒ *y* − 4*x* = −1

If (*x*, *y*) is a point on this line, then it must satisfy the above equation.

Similarly, if (*x* + 1, *y* + 4) is a point on this line, then it must also satisfy the above equation.

Putting the coordinates (*x* + 1, *y* + 4) in the above equation:

*y* + 4 − 4(*x* + 1) = −1

⇒ *y* + 4 − 4*x* − 4 = −1

⇒ *y* − 4*x* = −1

This equation is identical to the original equation.

So, the point (*x* + 1, *y* + 4) also lie on the line *y* − 4*x* = −1 as the point (*x*, *y*).

#### Page No 190:

#### Question 3:

What is the point at which the line 2*x* + 4*y* − 1 = 0 cuts the *x*-axis? What about the *y*-axis?

#### Answer:

If a line cuts the *x*-axis, then the *y*-coordinate of that point should be zero.

For the equation, 2*x* + 4*y *− 1= 0, putting *y* = 0, we get:

2*x* + 4(0)* *− 1= 0

⇒ 2*x* − 1= 0

⇒ *x* =

Thus, the point at which the line 2*x* + 4*y** *− 1= 0 cuts the *x*-axis is

Similarly, if a line cuts the *y*-axis, then the *x*-coordinate of that point should be zero.

For the equation, 2*x* + 4*y *− 1= 0, putting *x* = 0, we get:

2(0) + 4*y *− 1= 0

⇒ 4*y* − 1= 0

⇒ *y* =

Thus, the point at which the line 2*x* + 4*y** *− 1= 0 cuts the *y*-axis is

#### Page No 190:

#### Question 4:

Prove that the lines given by the equations 3*x* + 2*y* + 5 = 0 and 3*x* + 2*y* − 1 = 0 are parallel. At what points do they intersect the *x*-axis? And the *y*-axis?

#### Answer:

We know that two lines are parallel if they have equal slope.

Given: Equation of lines 3*x* + 2*y** *+ 5 = 0 and 3*x* + 2*y** *− 1 = 0

Taking the coordinates of the two points on the first line as (*x*_{1}, *y*_{1}) and (*x*_{2}, *y*_{2}):

3*x*_{1} + 2*y*_{1}* *+ 5 = 0

3*x*_{2} + 2*y*_{2}* *+ 5 = 0

(3*x*_{1} + 2*y*_{1}* *+ 5) − (3*x*_{2} + 2*y*_{2}* *+ 5) = 0

⇒ 3(*x*_{1}_{ }−* **x*_{2}) + 2(*y*_{1}_{ }− *y*_{2}) = 0

⇒

Thus, the slope of the first line is

Taking the coordinates of the two points on the second line as (*x*_{1}, *y*_{1}) and (*x*_{2}, *y*_{2}):

3*x*_{1} + 2*y*_{1}* *− 1 = 0

3*x*_{2} + 2*y*_{2}* *− 1 = 0

(3*x*_{1} + 2*y*_{1}* *− 1) − (3*x*_{2} + 2*y*_{2}* *− 1) = 0

⇒ 3(*x*_{1}_{ }−* **x*_{2}) + 2(*y*_{1}_{ }− *y*_{2}) = 0

⇒

Thus, the slope of the second line is

Thus, the given lines are parallel to each other.

If a line cuts the *x*-axis, then the *y*-coordinate of that point must be zero.

For the equation, 3*x* + 2*y** *+ 5 = 0, putting *y* = 0, we get:

3*x* + 2(0)* *+ 5 = 0

⇒ 3*x* + 5= 0

⇒ *x* =

Therefore, the point at which the line 3*x* + 2*y *+ 5 = 0 cuts the *x*-axis is

Similarly, for the equation 3*x* + 2*y** *− 1= 0, putting *y* = 0, we get:

3*x* + 2(0)* *− 1 = 0

⇒ 3*x* − 1 = 0

⇒ *x* =

Therefore, the point at which the line 3*x* + 2*y *− 1 = 0 cuts the *x*-axis is

Similarly, if a line cuts the *y*-axis then the *x*-coordinate of that point must be zero.

For the equation, 3*x* + 2*y *+ 5 = 0, putting *x* = 0, we get:

3(0) + 2*y *+ 5 = 0

⇒ 2*y* + 5 = 0

⇒ *y* =

Therefore, the point at which the line 3*x* + 2*y** *+ 5 = 0 cuts the *y*-axis is

For the equation 3*x* + 2*y** *− 1 = 0, putting *x* = 0, we get:

3(0) + 2*y *− 1 = 0

⇒ 2*y* − 1= 0

⇒ *y* =

Therefore, the point at which the line 3*x* + 2*y** *− 1 = 0 cuts the *y*-axis is

#### Page No 190:

#### Question 5:

At what point do the lines of equations 3*x* + 2*y* + 5 = 0 and 2*x* − 3*y** *− 1 = 0 intersects each other? Find one more point on each of these lines. Prove that these lines are perpendicular.

#### Answer:

Intersection point of two lines is given by the coordinates obtained on solving the two linear equations.

Given equations of lines are 3*x* + 2*y* + 5 = 0 and 2*x* − 3*y* − 1 = 0.

On solving, we get* x* = −1, *y* = −1

Therefore, the intersection point of these lines is (−1, −1).

Let the point (*x*_{1}, *y*_{1}) lie on the line, 3*x* + 2*y* + 5 = 0.

⇒ 3*x*_{1} + 2*y*_{1} + 5 = 0

Any point on this line can be found out by assuming *x* to have any value, say 1 and putting it in the above equation.

⇒ 3(1) + 2*y*_{1} + 5 = 0

⇒ 2*y*_{1} = −8

⇒ *y*_{1} = −4

Thus, the point (1, −4) lie on the line 3*x* + 2*y* + 5 = 0.

Similarly, let the point (*x*_{2}, *y*_{2}) lie on the line 2*x* − 3*y* − 1 = 0.

⇒ 2*x*_{1} − 3*y*_{1} − 1 = 0

Any point on this line can be found out by assuming *x* to have any value, say 2 and putting it in the above equation.

⇒ 2(2) − 3*y*_{1} − 1 = 0

⇒ −3*y*_{1} = −3

⇒ *y*_{1} = 1

Thus, the point (2, 1) lie on the line 2*x* − 3*y* − 1 = 0.

Now, let the slope of the line 3*x* + 2*y* + 5 = 0 be *p* and the slope of 2*x* − 3*y* − 1 = 0 be *q*.

We have just now found the points lying on these lines.

Points (−1, −1) and (1, −4) lie on the line 3*x* + 2*y* + 5 = 0.

∴ *p* =

Similarly, points (−1, −1) and (2, 1) lie on the line 2*x* − 3*y* − 1 = 0.

∴ *q* =

For two lines to be perpendicular, the product of their slopes should be equal to −1, i.e., *pq *_{ }= −1

*pq *= × = −1

Thus, the given lines are perpendicular.

View NCERT Solutions for all chapters of Class 10