Rs Aggarwal 2015 Solutions for Class 10 Math Chapter 16 Coordinate Geometry are provided here with simple step-by-step explanations. These solutions for Coordinate Geometry are extremely popular among Class 10 students for Math Coordinate Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2015 Book of Class 10 Math Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2015 Solutions. All Rs Aggarwal 2015 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### Question 1:

Find the distance between the points:

(i) A(9, 3) and B(15, 11)
(ii) A(7, −4) and B(−5, 1)
(iii) A(−6, −4) and B(9, −12)
(iv) A(1, −3) and B(4, −6)
(v) P(a + b, ab) and Q(ab, a + b)
(vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)

(i) A(9, 3) and B(15, 11)
The given points are A(9, 3) and B(15, 11).
Then (x1 = 9, y1 = 3) and (x2 = 15, y2 = 11)

(ii) A(7, −4) and B(−5, 1)
The given points are A(7, −4) and B(−5, 1).
Then (x1= 7, y1 = −4) and (x2 = −5, y2 = 1)

(iii) A(−6, −4) and B(9, −12)
The given points are A(−6, −4) and B(9, −12).
Then (x1 = −6, y1 = −4) and (x2 = 9, y2 = −12)

(iv) A(1, −3) and B(4, −6)
The given points are A(1, −3) and B(4, −6).
Then (x1 = 1, y1 = −3) and (x2 = 4, y2 = −6)

(v) P(a + b, ab) and Q(ab, a + b)
The given points are P(a + b, a − b) and Q(a − b, a + b).
Then (x1 = a + b, y1 = a − b) and (x2 = a − b, y2 = a + b)

(vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)
The given points are P(a sin α, a cos α) and Q(a cos α, −a sin α).
Then (x1 = a sin α, y1 = a cos α) and (x2 = a cos α, y2 = −a sin α)

#### Question 2:

Find the distance of each of the following points from the origin:

(i) A(5, −12)
(ii) B(−5, 5)
(iii) C(−, −6)4

(i) A(5, −12)
Let O(0, 0) be the origin.

(ii) B(−5, 5)
Let O(0, 0) be the origin.

(iii) C(−4, −6)
Let O(0,0) be the origin.

#### Question 3:

Find all possible values of a for which the distance between the points A(a, −1) and B(5, 3) is 5 units.

Given AB = 5 units
Therefore, (AB)2 = 25 units

Therefore, a = 2 or 8.

#### Question 4:

Find the ordinate of a point whose abscissa is 10 and which is at a distance of 10 units from the points P(2, −3).

Let Q be the other point whose coordinates are Q(10, y).
Given, PQ = 10 units
Therefore, (PQ)2 = 100

Therefore, y = 3 or −9.

#### Question 5:

If A(6, −1), B(1, 3) and C(k, 8) are three points such that AB = BC, find the value of k.

The given points are A(6, −1), B(1, 3) and C(k, 8).
Also, it is given that AB = BC.
Therefore, (AB)2 = (BC)2

Therefore, the value of k will be either 5 or −3.

#### Question 6:

If the point A(a, 2) is equidistant from the points B(8, −2) and C(2, −2), find the value of a.

The given points are A(a, 2), B(8, −2) and C(2, −2).
Also, it is given that AB = AC.
Therefore, (AB)2 = (AC)2

Therefore, the value of a is 5.

#### Question 7:

Find the point on the x-axis that is equidistant from the points (−2, 5) and (−2, 9).

Let the given points be A(−2, 5) and B(−2, 9) and the required point be P(x,0). Then, AP = PB.
That is, (AP)2 = (PB)2

Disclaimer : As it is never possible that 0 will become equal to 56
Therefore, we can say that there is no point on the x-axis which is equidistance from A(-2,5) and B(-2,9)

#### Question 8:

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

Let the given points be A(5, −2) and B(−3, 2) and the required point be P(0, y). Then, AP = BP.
Also, (AP)2 = (BP)2

So, the point is (0, −2).

#### Question 9:

If the points A(4, 3) and B(x, 5) lie on a circle with centre O(2, 3), find the value of x.

Given, the points A(4, 3) and B(x, 5) lie on a circle with centre O(2, 3).
Then OA = OB
Also (OA)2 = (OB)2

Therefore, x = 2.

#### Question 10:

If P(x, y) is a point equidistant from the points A(6, −1) and B(2, 3), show that xy = 3.

The given points are A(6, −1) and B(2, 3). The point P(x, y) is equidistant from the points A and B. So, PA = PB.
Also, (PA)2 = (PB)2

Hence proved.

#### Question 11:

Find points on the x-axis, each of which is at a distance of 10 units from the point A(11, −8).

A(11,−8) is the given point. Let P(x,0) be the required point on the x-axis; then, AP = 10 units.
Also (AP)2 = (10)2 = 100

Therefore, the coordinates of the point are (17, 0) or (5, 0). It is 10 units away from the point A(11, −8).

#### Question 12:

Find the coordinates of the point equidistant from the three points A(5, 3), B(5, −5) and C(1, −5).

Let the required point be P(x, y). Then AP = BP = CP
That is, (AP)2=(BP)2=(CP)2
This means (AP)2=(BP)

Hence, the required point is (3, −1).

#### Question 13:

Using the distance formula, show that the points (1, −1), (5, 2) and (9, 5) are collinear.

Let A(1, −1), B (5, 2) and C(9, 5) be the given points. Then:

Hence, the given points are collinear.

#### Question 14:

Show that the following points are collinear:

(i) A(6, 9) B(0, 1) and C(−6, −7)
(ii) A(−1, −1) B(2, 3) and C(8, 11)
(iii) P(1, 1) Q(−2, 7) and R(3, −3)
(iv) P(2, 0) Q(11, 6) and R(−4, −4)

(i) A(6, 9) B(0, 1) and C(−6, −7)

Hence, the points A, B and C are collinear.

(ii) A(−1, −1) B(2, 3) and C(8, 11)

Hence, the points A, B and C are collinear.

(iii) P(1, 1) Q(−2, 7) and R(3, −3)

Hence, the points Q, P and R are collinear.

(iv) P(2, 0) Q(11, 6) and R(−4, −4)

Hence, the points Q, P and R are collinear.

#### Question 15:

Show that the points A(3, 0), B(6, 4) and C(−1, 3) are the vertices of a right-angled triangle. Also, prove that these are the vertices of an isosceles triangle.

A(3, 0), B (6, 4) and C(−1, 3) are the given points. Then:

Thus, AB = BC = 5 units
Also, (AB)2+(AC)2 = (5)2+(5)2 = 50
and (BC)2 = (5$\sqrt{2}$)2 = 25$×$2 = 50
Thus, (AB)2+(AC)2 = (BC)2
This show that $∆ABC$ is right- angled at A.
This proves that the the points A(3, 0), B(6, 4) and C(−1, 3) are the vertices of an isosceles right- angled triangle.

#### Question 16:

Prove that the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right triangle.

The given points are A(7, 10), B(−2, 5) and C(3, −4).

Since, AB and BC are equal, they form the vertices of an isosceles triangle.
Also, (AB)2+(BC)2
and (AC)2 = ${\left(\sqrt{212}\right)}^{2}$ = 212
Thus, (AB)2+(BC)2 = (AC)2
This show that $∆ABC$ is right- angled at B.
Therefore, the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right-angled triangle.

#### Question 17:

Show that the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle. Calculate its area.

Let the given points be A(−5, 6) B(3, 0) and C(9, 8).

Also, (AB)2+(BC)2
and (AC)2 = ${\left(10\sqrt{2}\right)}^{2}=200$
Thus, (AB)2+(BC)2 = (AC)2
This show that $∆ABC$ is right- angled at B.
Therefore, the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle.
Also,

#### Question 18:

Show that the points O(0, 0) A(3, $\sqrt{3}$) and B(3, −$\sqrt{3}$) are the vertices of an equilateral triangle. Find the area of this triangle.

The given points are O(0, 0) A(3, $\sqrt{3}$) and B(3, − $\sqrt{3}$).

Thus, t
he points O(0, 0) A(3, $\sqrt{3}$)and B(3, − $\sqrt{3}$) are the vertices of an equilateral triangle.
Also, the area of the triangle OAB = $\frac{\sqrt{3}}{4}×{\left(\mathrm{side}\right)}^{2}$

#### Question 19:

Show that the points A(2, 1), B(5, 2), C(6, 4) and D(3, 3) are the angular points of a parallelogram. Is this figure a rectangle?

The given points are A(2, 1), B(5, 2), C(6, 4) and D(3, 3).

But diagonal AC is not equal to diagonal BD.
Hence, the given points do not form a rectangle.

#### Question 20:

Show that the following points are the vertices of a rectangle.

(i) A(0, −4), B(6, 2) C(3, 5) and D(−3, −1)
(ii) A(2, −2), B(14, 10) C(11, 13) and D(−1, 1)
(iii) A(−4, −1), B(−2, −4) C(−3, 2) and D(0, −1)

(i) The given points are A(0, −4), B(6, 2) C(3, 5) and D(−3, −1).

Also, diagonal AC = diagonal BD.
Hence, the given points form a rectangle.

(ii) The given points are A(2, −2), B(14, 10) C(11, 13) and D(−1, 1).

Also, diagonal AC = diagonal BD
Hence, the given points form a rectangle.

(iii) The given points are A(−4, −1), B(−2, −4) C(4, 0) and D(2, 3).

Also, diagonal AC = diagonal BD
Hence, the given points form a rectangle.

#### Question 21:

Show that the following points are the vertices of a square.

(i) A(6, 2), B(2, 1), C(1, 5) and D(5, 6)
(ii) P(0, −2), Q(3, 1), R(0, 4) and S(−3, 1)
(iii) A(3, 2), B(0, 5), C(−3, 2) and D(0, −1)

(i) The given points are A(6, 2), B(2, 1), C(1, 5) and D(5, 6).

Therefore, the given points form a square.

(ii) The given points are P(0, −2), Q(3, 1), R(0, 4) and S(−3, 1).

Therefore, the given points form a square.

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

Therefore, the given points form a square.

#### Question 22:

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

The given points are A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4).

Therefore, ABCD is a quadrilateral with equal sides and unequal diagonals.
Hence, ABCD is a rhombus.

#### Question 1:

Find the coordinates of the point that divides the join of A(−1, 7) and B(4, −3) in the ratio 2 : 3.

The end points of AB are A(−1, 7) and B(4, −3).
Therefore, (x1 = −1, y1 = 7) and (x2 = 4, y2 = −3)
Also, m = 2 and n = 3
Let the required point be P(x, y).
By section formula, we get:

Hence, the coordinates of the required point are (1, 3).

#### Question 2:

Find the coordinates of the points that divide the join of A(−5, 11) and B(4, −7) in the ratio 7 : 2.

The end points of AB are A(−5, 11) and B(4, −7).
Therefore, (x1 = −5, y1 = 11) and (x2 = 4, y2 = −7).
Also, m = 7 and n = 2
Let the required point be P(x, y).
By section formula, we have:

Hence, the required point is P(2, −3).

#### Question 3:

Find the coordinates of the points of trisection of the line segment AB, whose end points are A(2, 1) and B(5, −8).

Let P and Q be the points of trisection of AB.
Then P divides AB in the ratio 1:2.
So the coordinates of P are

Therefore, the coordinates of P are (3, −2).
Also, Q divides AB in the ratio 2:1.
So, the coordinates of Q are

Therefore, the coordinates of Q are (4, −5).
Hence, the points are (3, −2) and (4, −5).

#### Question 4:

Find the coordinates of the points that divide the join of A(−4, 0) and B(0, 6) in three equal parts.

Let P and Q be the points of trisection of AB.
Then, P divides AB in the ratio 1:2.
So, the coordinates of P are

Also Q divides AB in the ratio 2:1
So the coordinates of Q are

Hence, the points are .

#### Question 5:

The line segment joining the points A(3, −4) and B(1, 2) is trisected at the points P(p, −2) and $Q\left(\frac{5}{3},q\right)$. Find the values of p and q.

Let P and Q be the points of trisection of AB.
Then, P divides AB in the ratio 1:2.
So, the coordinates of P are

Hence, the coordinates of P are ($\frac{7}{3}$, −2).
But (p, −2) are the coordinates of P.
So, $p=\frac{7}{3}$
Also, Q divides the line AB in the ratio 2:1.
So, the coordinates of Q are

But the given coordinates of
So, q = 0
Thus, $p=\frac{7}{3}$ and $q=0$53

#### Question 6:

The line segment joining the points A(4, −5) and B(4, 5) is divided by the point P such that $\frac{AP}{AB}=\frac{2}{5}$. Find the coordinates of P.

The given points are A(4, −5) and B(4, 5).
P divides AB, such that .

#### Question 7:

Find the coordinates of the midpoints of the line segment joining:

(i) A(3, 0) and B(−5, 4)
(ii) P(−11, −8) and Q(8, −2)

(i) The given points are A(3, 0) and B(−5, 4).
Let (x, y) be the mid point of AB. Then:

Therefore, (−1, 2) are the coordinates of mid point of AB.

(ii) The given points are P(−11, −8) and Q(8, −2).
Let (x, y) be the mid point of PQ. Then:

Therefore, are the coordinates of midpoint of PQ.

#### Question 8:

If (2, p) is the midpoint of the line segment joining the points A(6, −5) and B(−2, 11), find the value of p.

The given points are A(6, −5) and B(−2, 11).
Let (x, y) be the mid point of AB. Then:

So, the midpoint of AB is (2, 3).
But it is given that the midpoint of AB is (2, p).
Therefore, the value of p = 3.

#### Question 9:

The midpoint of the line segment joining A(2a, 4) and B(−2, 3b) is C(1, 2a + 1). Find the values of a and b.

The points are A(2a, 4) and B(−2, 3b).
Let C(1, 2a + 1) be the mid point of AB. Then:

#### Question 10:

Points P, Q and R, in that order, divide a line segment joining A(1, 6) and B(5, −2) in four equal parts. Find the coordinates of P, Q and R.

The given points are A(1, 6) and B(5, −2).
Then, P(x, y) is a point that divides the line AB in the ratio 1:3.
By the section formula:

Therefore, the coordinates of point P are (2, 4).
Let Q be the mid point of AB.
Then, Q(x, y):

Therefore, the coordinates of Q are (3, 2).
Let R (x, y) be a point that divides AB in the ratio 3:1.
Then, by the section formula:

Therefore, the coordinates of R are (4, 0).
Hence, the coordinates of point P, Q and R are (2, 4), (3, 2) and (4, 0) respectively.

#### Question 11:

The line segment joining A(−2, 9) and B(6, 3) is a diameter of a circle with centre C. Find the coordinates of C.

The given points are A(−2, 9) and B(6, 3).
Then, C(x, y) is the midpoint of AB.

Therefore, the coordinates of point C are (2, 6).

#### Question 12:

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

C(2, −3) is the centre of the given circle. Let A(a, b) and B(1, 4) be the two end-points of the given diameter AB. Then, the coordinates of C are

Therefore, the coordinates of point A are (3, -10).

#### Question 13:

In what ratio does the point P(2, 5) divide the join of A(8, 2) and B(−6, 9)?

Let the point P(2, 5) divide AB in the ratio k : 1.
Then, by section formula, the coordinates of P are

Therefore, the point P(2, 5) divides AB in the ratio 3 : 4.

#### Question 14:

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

Let the point P(−6, a) divide the line AB in the ratio k : 1.
Then, by the section formula:

#### Question 15:

Find 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.

Let the point P(m, 6) divide the line AB in the ratio k : 1.
Then, by the section formula:

#### Question 16:

Find the ratio in which the point (−3, k) divides the join of A(−5, −4) and B(−2, 3). Also, find the value of k.

Let the point P(−3, k) divide the line AB in the ratio s : 1.
Then, by the section formula:

#### Question 17:

In what ratio is the line segment joining A(2, −3) and B(5, 6) divided by the x-axis? Also, find the coordinates of the point of division.

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are

But P lies on the x-axis; so, its ordinate is 0.

Therefore, the required ratio is $\frac{1}{2}$ : 1, which is same as 1 : 2.
Thus, the x-axis divides the line AB in the ratio 1 : 2 at the point P.
Applying k = $\frac{1}{2}$, we get the coordinates of point :

Hence, the point of intersection of AB and the x-axis is P(3, 0).

#### Question 18:

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

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are
$P\left(\frac{3k-2}{k+1},\frac{7k-3}{k+1}\right)$
But P lies on the y-axis; so, its abscissa is 0.

Therefore, the required ratio is $\frac{2}{3}$ : 1, which is same as 2 : 3.
Thus, the x-axis divides the line AB in the ratio 2:3 at the point P.
Applying k=$\frac{2}{3}$, we get the coordinates of point P:

Hence, the point of intersection of AB and the x-axis is P(0, 1).

#### Question 19:

In what ratio does the line xy − 2 = 0 divide the line segment joining the points A(3, −1) and B(8, 9)?

Let the line xy − 2 = 0 divide the line segment joining the points A(3, −1) and B(8, 9) in the ratio k : 1 at P.
Then, the coordinates of P are

Since, P lies on the line xy − 2 = 0, we have:

$\left(\frac{8k+3}{k+1}\right)-\left(\frac{9k-1}{k+1}\right)-2=0\phantom{\rule{0ex}{0ex}}⇒8k+3-9k+1-2k-2=0\phantom{\rule{0ex}{0ex}}⇒8k-9k-2k+3+1-2=0\phantom{\rule{0ex}{0ex}}⇒-3k+2=0\phantom{\rule{0ex}{0ex}}⇒-3k=-2\phantom{\rule{0ex}{0ex}}⇒k=\frac{2}{3}$
So, the required ratio is $\frac{2}{3}$ : 1, which is equal to 2 : 3.

#### Question 20:

Find the lengths of the median of ∆ABC whose vertices are A(0, −1), B(2, 1) and C(0, 3).

The vertices of ∆ABC are A(0, −1), B(2, 1) and C(0, 3).
Let AD, BE and CF be the medians of ∆ABC.

Let D be the midpoint of BC. So, the coordinates of D are

Let E be the midpoint of AC. So, the coordinates of E are

Let F be the midpoint of AB. So, the coordinates of F are

Therefore, the lengths of the medians: AD$\sqrt{10}$ units, BE = 2 units and CF = $\sqrt{10}$ units

#### Question 21:

Find the centroid of ∆ABC whose vertices are A(−1, 0), B(5, −2) and C(8, 2).

Here, (x1 = −1, y1 = 0), (x2 = 5, y2 = −2) and (x3 = 8, y3 = 2).
Let G(x, y) be the centroid of the ∆ABC. Then,

Hence, the centroid of ∆ABC is G(4, 0).

#### Question 22:

If G(−2, 1) is the centroid of a ∆ABC and two of its vertices are A(1, −6) and B(−5, 2), find the third vertex of the triangle.

Two vertices of ∆ABC are A(1, −6) and B(−5, 2). Let the third vertex be C(a, b).
Then the coordinates of its centroid are

But it is given that G(−2, 1) is the centroid. Therefore,

Therefore, the third vertex of ∆ABC is C(−2, 7).

#### Question 23:

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

Two vertices of ∆ABC are B(−3,1) and C(0, −2). Let the third vertex be A(a, b).
Then, the coordinates of its centroid are

But it is given that the centroid is at the origin, that is G(0, 0). Therefore,

Therefore, the third vertex of ∆ABC is A(3, 1).

#### Question 24:

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

The points are A(3, 1), B(0, −2), C(1, 1) and D(4, 4).
Join AC and BD, intersecting at O.

We know that the diagonals of a parallelogram bisect each other.

Thus, the diagonals AC and BD have the same midpoint.
Therefore, ABCD is a parallelogram.

#### Question 25:

If the points P(a, −11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a parallelogram PQRS, find the values of a and b.

The points are P(a, −11), Q(5, b), R(2, 15) and S(1, 1).

Join PR and QS, intersecting at O.
We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of PR as well as QS.

#### Question 26:

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

Let A(1, −2), B(3, 6) and C(5, 10) be the three vertices of a parallelogram ABCD and the fourth vertex be D(a, b).
Join AC and BD intersecting at O.

We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of AC as well as BD.

Therefore, the fourth vertex is D(3, 2).

#### Question 1:

Find the area of ∆ABC whose vertices are:

(i) A(1, 2), B(−2, 3) and C(−3, −4)
(ii) A(−5, 7), B(−4, −5) and C(4, 5)
(iii) A(3, 8), B(−4, 2) and C(5, −1)
(iv) A(10, −6), B(2, 5) and C(−1, 3)

(i) A(1, 2), B(−2, 3) and C(−3, −4) are the vertices of ∆ABC. Then,
(x1 = 1, y1 = 2), (x2 = −2, y2 = 3) and (x3 = −3, y3 = -4)

(ii) A(−5, 7), B(−4, −5) and C(4, 5) are the vertices of ∆ABC. Then,
(x1 = −5, y1 = 7), (x2 = −4, y2 = −5) and (x3 = 4, y3 = 5)

(iii) A(3, 8), B(−4, 2) and C(5, −1) are the vertices of ∆ABC. Then,
(x1 = 3, y1 = 8), (x2 = −4, y2 = 2) and (x3 = 5, y3 = −1)

(iv) A(10, −6), B(2, 5) and C(−1, −3) are the vertices of ∆ABC. Then,
(x1 = 10, y1 = −6), (x2 = 2, y2 = 5) and (x3 = −1, y3 = 3)

#### Question 2:

(i) Find the area of a quadrilateral ABCD whose vertices are
A(−4, −2), B(−3, −5), C(3, −2) and D(2, 3)

(ii) Find the area of the quadrilateral ABCD whose vertices are
A(0, 0), B(6, 0), C(4, 3) and D(0, 3)

(iii) Find the area of the quadrilateral ABCD whose vertices are
A(1, 0), B(5, 3), C(2, 7) and D(−2, 4).

(i) The vertices are A(−4, −2), B(−3, −5), C(3, −2) and D(2, 3).
Join AC. Then,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

Now,

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of quadrilateral ABCD = 10.5 sq. units + 17.5 sq. units = 28 sq. units

(ii) The vertices are A(0, 0), B(6, 0), C(4, 3) and D(0, 3).
Join AC. Then,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

Now,

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of quadrilateral ABCD = 9 sq. units + 6 sq. units = 15 sq. units

(iii) The vertices are A(1, 0), B(5, 3), C(2, 7) and D(−2, 4).
Join AC. Then,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

Now,

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of quadrilateral ABCD = 12.5 sq. units + 12.5 sq. units = 25 sq. units

#### Question 3:

Show that the following points are collinear:

(i) A(0, 1), B(1, 2) and C(−2, −1)
(ii) A(−5, 1), B(5, 5) and C(10, 7)
(iii) P(a, b + c), Q(b, c + a) and R(c, a + b)

(i) A(0, 1), B(1, 2) and C(−2, −1) are the given points. Then:
(x1 = 0, y1 = 1), (x2 = 1, y2 = 2) and (x3 =−2, y3 = −1)
$\therefore {x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=0\left(2-\left(-1\right)\right)+1\left(-1-1\right)+\left(-2\right)\left(1-2\right)\phantom{\rule{0ex}{0ex}}=0\left(3\right)+1\left(-2\right)-2\left(-1\right)\phantom{\rule{0ex}{0ex}}=0-2+2\phantom{\rule{0ex}{0ex}}=0$
Hence, the given points are collinear.

(ii) A(−5, 1), B(5, 5) and C(10, 7) are the given points. Then:
(x1 = −5, y1 = 1), (x2 = 5, y2 = 5) and (x3 = 10, y3 = 7)
$\therefore {x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=\left(-5\right)\left(5-7\right)+5\left(7-1\right)+10\left(1-5\right)\phantom{\rule{0ex}{0ex}}=-5\left(-2\right)+5\left(6\right)+10\left(-4\right)\phantom{\rule{0ex}{0ex}}=10+30-40\phantom{\rule{0ex}{0ex}}=0$
Hence, the given points are collinear.

(iii) P(a, b + c), Q(b, c + a) and R(c, a + b) are the given points. Then:
(x1 = a, y1 = b + c), (x2 = b, y2 = c + a) and (x3 = c, y3 = a + b)
$\therefore {x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=a\left(c+a-a-b\right)+b\left(a+b-b-c\right)+c\left(b+c-c-a\right)\phantom{\rule{0ex}{0ex}}=a\left(c-b\right)+b\left(a-c\right)+c\left(b-a\right)\phantom{\rule{0ex}{0ex}}=ac-ab+ba-bc+cb-ca\phantom{\rule{0ex}{0ex}}=0$
Hence, the given points are collinear.

#### Question 4:

Find the values of p for which the given points are collinear:

(i) A(−1, 3) B(2, p) and C(5, −1)
(ii) A(3, 2) B(4, p) and C(5, 3)
(iii) A(−3, 9) B(2, p) and C(4, −5)

(i) A(−1, 3), B(2, p) and C(5, −1) are the given points. Then:
(x1 = −1, y1 = 3), (x2 = 2, y2 = p) and (x3 = 5, y3 = −1)
It is given that points A, B and C are collinear.

Therefore, the value of p is 1.

(ii) A(3, 2) B(4, p) and C(5, 3) are the given points. Then:
(x1 = 3, y1 = 2), (x2 = 4, y2 = p) and (x3 = 5, y3 = 3)
It is given that points A, B and C are collinear.

Therefore, the value of p is $\frac{5}{2}$.

(iii) A(−3, 9) B(2, p) and C(4, −5) are the given points. Then:
(x1 = −3, y1 = 9), (x2 = 2, y2 = p) and (x3 = 4, y3 = −5)
It is given that points A, B and C are collinear.

Therefore, the value of p is −1.

#### Question 5:

For what value of x are the points A(−3, 12), B(7, 6) and C(x, 9) collinear?

A(−3, 12), B(7, 6) and C(x, 9) are the given points. Then:
(x1 = −3, y1 = 12), (x2 = 7, y2 = 6) and (x3 = x, y3 = 9)
It is given that the points A, B and C are collinear. Therefore,

Therefore, when x= 2, the given points are collinear.

#### Question 6:

For what value of y, are the points P(1, 4), Q(3, y) and R(−3, 16) collinear?

P(1, 4), Q(3, y) and R(−3, 16) are the given points. Then:
(x1 = 1, y1 = 4), (x2 = 3, y2 = y) and (x3 = −3, y3 = 16)
It is given that the points P, Q and R are collinear.
Therefore,

When y = −2, the given points are collinear.

#### Question 7:

If the points A(x, y), B(−5, 7) and C(−4, 5) are collinear, then show that 2x + y + 3 = 0.

The given points are A(x, y), B(−5, 7) and C(−4, 5).
So, (x1 = x, y1 = y), (x2 = −5, y2 = 7) and (x3 = −4, y3 = 5).
If the given points are collinear, then:

#### Question 8:

A(−4, −2), B(−3, −5) C(3, −2) and D(2, k) are the vertices of a quadrilateral ABCD. Find the value of k, if the area of the quadrilateral is 28 sq units.

Let A(−4, −2), B(−3, −5), C(3, −2) and D(2, k) be the vertices of quadrilateral ABCD.
Join AC. Then, area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD.

To find the value of k, let us consider the area of triangle ACD.

Therefore, the value of k = 3.

#### Question 9:

For ∆ABC, whose vertices are A(4, −6), B(3, −2) and C(5, 2), verify that a median of the triangle divides it into two triangles of equal areas.

To prove : Area of triangle ABD = Area of triangle ACD
Given ∆ABC, whose vertices are A(4, −6), B(3, −2) and C(5, 2).
Join AD. Let AD be the median that divides the triangle into two triangles.

So, D is the midpoint of BC.
Therefore, the coordinates of D are

Now,

and

Therefore, area of triangle ABD = Area of triangle ACD
Hence, the median of ∆ABC divides it into two triangles of equal areas.

#### Question 10:

Find a relation between x and y if the points A(2, 1), B(x, y) and C(7, 5) are collinear.

Consider the points A(2, 1), B(x, y) and C(7, 5) .
Here, (x1 = 2, y1 = 1), (x2 = x, y2 = y) and (x3 = 7, y3 = 5).
It is given that the points are collinear. So,

#### Question 11:

Prove that the points A(a, 0), B(0, b) and C(1, 1) are collinear if $\left(\frac{1}{a}+\frac{1}{b}\right)=1.$

Consider the points A(a, 0), B(0, b) and C(1, 1).
Here, (x1 = a, y1 = 0), (x2 = 0, y2 = b) and (x3 = 1, y3 = 1).
It is given that the points are collinear. So,

Therefore, the given points are collinear if $\left(\frac{1}{a}+\frac{1}{b}\right)$ = 1.

#### Question 1:

Find the distance between the points

The given points are .
Then,
Therefore,

#### Question 2:

Find the value of a, so that the point (3, a) lies on the line 2x − 3y = 5.

$⇒6-3a=5\phantom{\rule{0ex}{0ex}}⇒3a=1\phantom{\rule{0ex}{0ex}}⇒a=\frac{1}{3}$
.

#### Question 3:

If the points A(4, 3) and B(x, 5) lie on the circle with centre O(2, 3), find the value of x.

.

#### Question 4:

If P(x, y) is equidistant from the points A(7, 1) and B(3, 5), find the relation between x and y.

Let the point  be equidistant from the points A(7, 1) and B(3, 5).
Then,
$PA=PB\phantom{\rule{0ex}{0ex}}⇒P{A}^{2}=P{B}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-7\right)}^{2}+{\left(y-1\right)}^{2}={\left(x-3\right)}^{2}+{\left(y-5\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-14x-2y+50={x}^{2}+{y}^{2}-6x-10y+34\phantom{\rule{0ex}{0ex}}⇒8x-8y=16\phantom{\rule{0ex}{0ex}}⇒x-y=2$

#### Question 5:

If the centroid of ∆ABC, which has vertices A(a, b), B(b, c) and C(c, a), is the origin, find the value of (a + b + c).

The given points are A(a, b), B(b, c) and C(c, a).
Here,

Let the centroid be (x, y).
Then,

But it is given that the centroid of the triangle is the origin.
Then, we have:
$\frac{a+b+c}{3}=0\phantom{\rule{0ex}{0ex}}⇒a+b+c=0$

#### Question 6:

Find the centroid of ∆ABC whose vertices are A(2, 2), B(−4, −4) and C(5, −8).

The given points are A(2, 2), B(−4, −4) and C(5, −8).
Here,
Let G(x, y) be the centroid of $∆ABC$. Then,

Hence, the centroid of  .

#### Question 7:

In what ratio does the point C(4, 5) divide the join of A(2, 3) and B(7, 8)?

Let the required ratio be .
Then, by section formula, the coordinates of C are
$C\left(\frac{7k+2}{k+1},\frac{8k+3}{k+1}\right)$
Therefore,

#### Question 8:

If the points A(2, 3) B(4, k) and C(6, −3) are collinear, find the value of k.

The given points are .
Here,
It is given that the points A, B and C​ are collinear. Then,
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒2\left(k+3\right)+4\left(-3-3\right)+6\left(3-k\right)=0\phantom{\rule{0ex}{0ex}}⇒2k+6-24+18-6k=0\phantom{\rule{0ex}{0ex}}⇒-4k=0\phantom{\rule{0ex}{0ex}}⇒k=0$

#### Question 1:

In which quadrant does the point (−3, 5) lie?

(a) I
(b) II
(c) III
(d) IV

(b) II
The point (−3, 5) lies in quadrant II, since the signs of the coordinates are (−, +).

#### Question 2:

In which quadrant does the point (2, −4) lie?

(a) I
(b) II
(c) III
(d) IV

(d) IV
The point (2, −4) lies in quadrant IV since the signs of the coordinates are (+, −).

#### Question 3:

The distance of the point P(3, 4) from the x-axis is

(a) 3 units
(b) 4 units
(c) 7 units
(d) 1 unit

(b) 4 units
The y-coordinate is the distance of the point from the x-axis.
Here, the y-coordinate is 4.

#### Question 4:

P is a point on x-axis at a distance of 3 units from y-axis, to its right. The coordinates of P are

(a) (3, 0)
(b) (0, 3)
(c) (3, 3)
(d) (−3, 3)

(a) (3, 0)

#### Question 5:

A is a point on the y-axis at a distance of 4 units from the x-axis and lying below the x-axis. The coordinates of A are

(a) (4, 0)
(b) (0, 4)
(c) (−4, 0)
(d) (0, −4)

(d) (0, −4)

#### Question 6:

The distance of the point P(4, −3) from the origin is

(a) 1 unit
(b) 3 units
(c) 5 units
(d) 7 units

(c) 5 units
The given point is P​(4, −3) and the origin is (0, 0).
Then,

#### Question 7:

The distance between the points A(2, −3) and B(2, 2) is

(a) 4 units
(b) 5 units
(c) 3 units
(d) 2 units

(b) 5 units
The given points are A(2, −3) and B(2, 2).
Then,
Therefore,

#### Question 8:

What point on the x-axis is equidistant from the points A(7, 6) and B(−3, 4)?

(a) (0, 4)
(b) (−4, 0)
(c) (3, 0)
(d) (0, 3)

(c) (3, 0)
Let P(x,0) be equidistant from the points A(7,6) and B(−3,4).
Then,

Hence,

#### Question 9:

The distance between the points A(0, 5) and B(−5, 0) is

(a) 5 units
(b) $2\sqrt{5}$ units
(c) $\sqrt{10}$ units
(d) $5\sqrt{2}$ units

(d) $5\sqrt{2}$ units
The given points are A(0, 5) and B(−5, 0).
Then,
Therefore,

#### Question 10:

A point P divides the join of A(5, −2) and B(9, 6) in the ratio 3 : 1. The coordinates of P are

(a) (4, 7)
(b) (8, 4)
(c) (12, 8)
(d) $\left(\frac{11}{2},5\right)$

(b) (8, 4)
The given points are A(5, −2) and B(9, 6)
Then,
Also, m ​= 3 and n = 1
Let the required point be P(x, y).
By section formula, we have

Hence, the coordinates of P are (8, 4).

#### Question 11:

In what ratio does the point P(1, 2) divide the join of A(−2, 1) and B(7, 4)?

(a) 1 : 2
(b) 2 : 1
(c) 3 : 2
(d) 2 : 3

(a) 1 : 2
Let the required ratio be k : 1.
Then, by section formula, the coordinates of P are
$P\left(\frac{7k-2}{k+1},\frac{4k+1}{k+1}\right)\phantom{\rule{0ex}{0ex}}$
Therefore,

$⇒k=\frac{1}{2}$ in each case
So, the required ratio is , which is same as 1 : 2.

#### Question 12:

The point which divides the line segment joining the points A(7, −6) and B(3, 4) in the ratio 1 : 2 lies in

The given points are and .
Then,
Also, m = 1 and n = 2
Let the required point be .
By section formula, we have

Hence, the coordinates of P are, which lies in the IV quadrant.

#### Question 13:

In what ratio does the x-axis divide the join of A(2, −3) and B(5, 6)?

(a) 2 : 3
(b) 3 : 5
(c) 1 : 2
(d) 2 : 1

(c) 1 : 2
Let AB be divided by the x axis in the ratio  at the point P.
Then, by section formula, the coordinates of are
$P\left(\frac{5k+2}{k+1},\frac{6k-3}{k+1}\right)$
Butlies on the x axis: so, its ordinate is 0.
$\frac{6k-3}{k+1}=0$
$⇒6k-3=0$
$⇒6k=3$
$⇒k=\frac{1}{2}$
Hence, the required ratio is , which is same as 1 : 2.

#### Question 14:

In what ratio does the y-axis divide the join of P(−4, 2) and Q(8, 3)?

(a) 3 : 1
(b) 1 : 3
(c) 2 : 1
(d) 1 : 2

(d) 1 : 2
Let AB be divided by the y axis in the ratio  at the point P.
Then, by section formula, the coordinates of are
$P\left(\frac{8k-4}{k+1},\frac{3k+2}{k+1}\right)$
But, P lies on the y axis; so, its abscissa is 0.
$⇒\frac{8k-4}{k+1}=0$
$⇒8k-4=0$
$⇒8k=4$
$⇒k=\frac{1}{2}$
Hence, the required ratio is , which is same as 1 : 2.

#### Question 15:

The midpoint of the line segment joining the points A(−2, 8) and B(−6, −4) is

(a) (4, 2)
(b) (−4, 2)
(c) (2, 6)
(d) (−4, −6)

(b) (−4, 2)
The given points are A(−2, 8) and B(−6, −4).
Then,
Let M(x,y) be the midpoint of AB. Then,

and

Hence, the required point is .

#### Question 16:

If P(−1, 1) is the midpoint of the line segment joining A(−3, b) and B(1, b + 4), then b = ?

(a) 1
(b) −1
(c) 2
(d) 0

(b) −1
The given points are A(−3, b) and B(1, b+4).
Then,
Therefore,

and

But the midpoint is .
Therefore,
$b+2=1\phantom{\rule{0ex}{0ex}}⇒b=-1$

#### Question 17:

If $P\left(\frac{a}{3},4\right)$ is the midpoint of the line segment joining A(−6, 5) and B(−2, 3), then a = ?

(a) −4
(b) −12
(c) 12
(d) −6

(b) −12
The given points are A(−6, 5) and B(−2, 3).
Then,
Therefore,

and

But the given midpoint is .
Therefore,
$-4=\frac{a}{3}\phantom{\rule{0ex}{0ex}}⇒a=-12$

#### Question 18:

If the distance between the points A(2, −2) and B(−1, x) is 5, then

(a) x = −3 or x = 4
(b) x = 3 or x = −4
(c) x = −6 or x = 2
(d) x = 6 or x = −2

(c) x = −6 or x = 2
The given points are A(2, −2) and B(−1, x) and AB = 5.
Then,
Therefore,

#### Question 19:

The line 2x + y −4 = 0 divides the line segment joining A(2, −2) and B(3, 7) in the ratio

(a) 2 : 5
(b) 2 : 9
(c) 2 : 7
(d) 2 : 3

(b) 2 : 9
Let the line​ divide the line segment in the ratio k : 1 at the point P.
Then, by section formula, the coordinates of P are
$P\left(\frac{3k+2}{k+1},\frac{7k-2}{k+1}\right)$
Since P lies on the line , we have:
$\frac{2\left(3k+2\right)}{k+1}+\frac{7k-2}{k+1}-4=0\phantom{\rule{0ex}{0ex}}⇒\left(6k+4\right)+\left(7k-2\right)-\left(4k+4\right)=0\phantom{\rule{0ex}{0ex}}⇒9k=2\phantom{\rule{0ex}{0ex}}⇒k=\frac{2}{9}$
Hence, the required ratio is , which is same as 2 : 9.

#### Question 20:

If A(4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC and AD is median, then the coordinates of D are

(a) $\left(\frac{5}{2},3\right)$
(b) $\left(5,\frac{7}{2}\right)$
(c) $\left(\frac{7}{2},\frac{9}{2}\right)$
(d) None of these

(c) $\left(\frac{7}{2},\frac{9}{2}\right)$
D is the midpoint of BC.
So, the coordinates of D are

#### Question 21:

If A(−1, 0), B(5, −2) and C(8, 2) are the vertices of a ∆ABC, then its centroid is

(a) (12, 0)
(b) (6, 0)
(c) (0, 6)
(d) (4, 0)

(d) (4, 0)
The given points are .
Here,
Let G(x, y) be the centroid of $∆ABC$. Then,

and

Hence, the centroid of $∆ABC$ is G(4, 0).

#### Question 22:

Two vertices of ∆ABC are A(−1, 4) and B(5, 2) and its centroid is G(0, −3). Then, the coordinates of C are

(a) (4, 3)
(b) (4, 15)
(c) (−4, −15)
(d) (−15, −4)

(c) (−4, −15)
Two vertices of .
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are

But it is given that the centroid is $G\left(0,-3\right)$.
Therefore,

Hence, the third vertex of .

#### Question 23:

If the distance between the points A(4, p) and B(1, 0) is 5, then

(a) p = 4 only
(b) p = −4 only
(c) p = ±4 only
(d) p = 0

(c) p = ±4 only
The given points are A(4, p) and B(1, 0) and AB = 5.
Then,
Therefore,
$AB=5\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}=5\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(1-4\right)}^{2}+{\left(0-p\right)}^{2}}=5\phantom{\rule{0ex}{0ex}}⇒{\left(-3\right)}^{2}+{\left(-p\right)}^{2}=25\phantom{\rule{0ex}{0ex}}⇒9+{p}^{2}=25\phantom{\rule{0ex}{0ex}}⇒{p}^{2}=16\phantom{\rule{0ex}{0ex}}⇒p=±\sqrt{16}\phantom{\rule{0ex}{0ex}}⇒p=±4\phantom{\rule{0ex}{0ex}}$

#### Question 24:

The three vertices of a parallelogram ABCD are A(−2, 3) B(6, 7) and C(8, 3). The fourth vertex D is

(a) (1, 0)
(b) (0, 1)
(c) (−1, 0)
(d) (0, −1)

(d) (0, −1)
Let the fourth vertex be D(a, b).
Join AC and BD.
Let AC and BD intersect at the point O.
We know that the diagonals of a parallelogram bisect each other.

Therefore,

Hence, the fourth vertex is D(0,−1).

#### Question 25:

The points A(−4, 0), B(4, 0) and C(0, 3) are the vertices of a triangle, which is

(a) isosceles
(b) equilateral
(c) scalene
(d) right-angled

(a) isosceles
Let A(−4, 0), B(4, 0) and C(0, 3) be the given points. Then,

BC = AC = 5 units
Therefore, $∆ABC$ is isosceles.

#### Question 26:

The points P(0, 6), Q(−5, 3) and R(3, 1) are the vertices of a triangle, which is

(a) equilateral
(b) isosceles
(c) scalene
(d) right-angled

(d) right-angled

Let P(0, 6), Q(−5, 3) and R(3, 1) be the given points. Then,

Therefore, ∆PQR is right-angled.

#### Question 27:

The points (a, a), (−a, a) and $\left(-\sqrt{3}a,\sqrt{3}a\right)$ form the vertices of

(a) an equilateral triangle
(b) a scalene triangle
(c) an isosceles triangle
(d) a right triangle

(a) an equilateral triangle
Let A(a, a), B(−a, −a) and  be the given points. Then,

AB = BC = AC
Hence, the points form the vertices of an equilateral triangle.

#### Question 28:

Three points A(1, −2) B(3, 4) and C(4, 7) form

(a) a straight line
(b) an equilateral triangle
(c) a right-angled triangle
(d) None of these

(a) a straight line
The given points are A(1, −2), B(3, 4) and C(4, 7).
Here, .
Therefore,

$=\frac{1}{2}\left[1\left(4-7\right)+3\left(7+2\right)+4\left(-2-4\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[-3+27-24\right]\phantom{\rule{0ex}{0ex}}=0$
Thus, the given points are collinear, i.e. they form a straight line.

#### Question 29:

If the points A(2, 3), B(5, k) and C(6, 7) are collinear, then

(a) k = 4
(b) k = 6
(c) $k=\frac{-3}{2}$
(d) $k=\frac{11}{4}$

(b) k = 6
The given points are A(2, 3), B(5, k) and C(6, 7).
Here,
Points A,B and C are collinear. Then,
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒2\left(k-7\right)+5\left(7-3\right)+6\left(3-k\right)=0\phantom{\rule{0ex}{0ex}}⇒2k-14+20+18-6k=0\phantom{\rule{0ex}{0ex}}⇒-4k=-24\phantom{\rule{0ex}{0ex}}⇒k=6$

#### Question 30:

If the points A(1, 2), O(0, 0) and C(a, b) are collinear, then

(a) a = b
(b) a = 2b
(c) 2a = b
(d) a + b = 0

(c) 2a = b
The given points are A(1, 2), O(0, 0) and C(a, b).
Here, .
Points A, O and C are collinear.
$⇒{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒1\left(0-b\right)+0\left(b-2\right)+a\left(2-0\right)=0\phantom{\rule{0ex}{0ex}}⇒-b+2a=0\phantom{\rule{0ex}{0ex}}⇒2a=b$

#### Question 31:

The area of ∆ABC with vertices A(a, b + c), B(b, c + a) and C(c, a + b) is

(a) (a + b + c)2
(b) a + b + c
(c) abc
(d) 0

(d) 0
The given points are .
Here,
Therefore,

$=\frac{1}{2}\left[a\left\{\left(c+a\right)-\left(a+b\right)\right\}+b\left\{\left(a+b\right)-\left(b+c\right)\right\}+c\left\{\left(b+c\right)-\left(c+a\right)\right\}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[a\left(c-b\right)+b\left(a-c\right)+c\left(b-a\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(ac-ab+ab-bc+bc-ac\right)\phantom{\rule{0ex}{0ex}}=0$

#### Question 32:

The point which lies on the perpendicular bisector of the line segment joining the points A(−2, −5) and B(2, 5) is

(a) (0, 0)
(b) (0, 2)
(c) (2, 0)
(d) (−2, 0)

(a) (0, 0)
The point which lies on the perpendicular bisector of AB is the midpoint of AB.
The given points are .
Here,

Hence, the required point is (0,0).

#### Question 33:

In the given figure, A(0, 2y) and B(2x, 0) are the end-points of line segment AB. The coordinates of point C, which is equidistant from the three vertices of ∆AOB, are

(a) (x , y)
(b) (y , x)
(c) $\left(\frac{x}{2},\frac{y}{2}\right)$
(d) $\left(\frac{y}{2},\frac{x}{2}\right)$

(a) (x , y)
The midpoint of AB is equidistant from the three vertices of $∆AOB.$
Therefore,

Hence, the coordinates of C are (x, y).

#### Question 34:

The points A(9, 0), B(9, 6), C(−9, 6) and D(−9, 0) are the vertices of a

(a) square
(b) rectangle
(c) rhombus
(d) trapezium

(b) rectangle
are the given vertices.
Then,

Therefore, we have:

Now, the diagonals are:

Therefore,
$A{C}^{2}=B{D}^{2}$
Hence, ABCD is a rectangle.

#### Question 35:

The area of ∆ABC with vertices A(3, 0), B(7, 0) and C(8, 4) is

(a) 14 sq units
(b) 28 sq units
(c) 8 sq units
(d) 6 sq units

(c) 8 sq units
The given points are .
Here,
Therefore,

#### Question 36:

AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). Length of each of its diagonal is

(a) 5 units
(b) 3 units
(c) $\sqrt{34}$ units
(d) 4 units

(c) $\sqrt{34}$ units
are the three vertices of a rectangle; let C be the fourth vertex.
Then, the length of the diagonal,

Since, the diagonals of rectangle are equal .
Hence, the length of its diagonals is .

#### Question 37:

A line intersects the y-axis and x-axis at the points P and Q, respectively. If (2, −5) is the midpoint of PQ, then the coordinates of P and Q, respectively, are

(a) (0, −5) and (2, 0)
(b) (0, −10) and (4, 0)
(c) (0, 10) and (−4, 0)
(d) (0, 4) and (−10, 0)

(b) (0, −10) and (4, 0)
Let the line PQ intersect y axis at P(0, a) and x axis at Q(b, 0).

But the midpoint of PQ is (2, −5).
Therefore,

Therefore,

Hence, the coordinates of P and Q are (0, −10) and (4, 0).

#### Question 38:

A circle drawn with origin as the centre passes through the point $A\left(\frac{13}{2},0\right)$. Which of the following points does not lie inside the circle?

(a) $\left(\frac{-3}{4},1\right)$
(b) $\left(2,\frac{7}{3}\right)$
(c) $\left(5,\frac{-1}{3}\right)$
(d) $\left(-6,\frac{5}{2}\right)$

(d) $\left(-6,\frac{5}{2}\right)$
Let the given point be  and the origin (0, 0). Then,
the radius of the circle,
$OA=\sqrt{\left(0-\frac{13}{2}\right)+{\left(0-0\right)}^{2}}$

Now, the distance of the point from the origin is

Now, the distance of the point from the origin is

Now, the distance of the point  from the origin is

Now, the distance of the point  from the origin is

Hence, it does not lie inside the circle.

#### Question 39:

The coordinates of one end point of a diameter AB of a circle are A(4, −1) and the coordinates of the centre of the circle are C(1, −3). Then, the coordinates of B are

(a) (2, −5)
(b) (−2, 5)
(c) (−2, −5)
(d) (2, 5)

(c) (−2, −5)
Let the coordinates of B be (a, b). Then,

But the midpoint C(1, −3) is given.
Therefore,

Hence, the coordinates of B are (−2, −5).

#### Question 40:

Match the following columns:

 Column I Column II (a) The coordinates of the point that divides the join of A(−1, 7) and B(4, −3) in the ratio 2 : 3 are ....... (p) (5, 6) (b) Two vertices of ∆ABC are A(6, 4) and B(−2, 2) and its centroid is G(3, 4). The coordinates of C are ....... (q) $\sqrt{10}$ (c) If the points A(4, 3) and B(x, 5) lie on a circle with the centre O(2, 3), then x = ....... . (r) (1, 3) (d) If A(0, −1), B(2, 1) and C(0, 3) are the vertices of a ∆ABC, then the length of median AD is (s) 2

(a) ⇒ (r)
(b) ⇒ (p)
(c) ⇒ (s)
(d) ⇒ (q)

(a) The given points are .
Then,
Also, m = 2 and n = 3
Let the required point be P(x, y).
By, section formula, we have:

Therefore, the coordinates of P are (1, 3).
Hence, $\left(a\right)⇒\left(r\right)$

(b) The two vertices of .
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are

But it is given that the centroid is G(3, 4).
Therefore,

Therefore, the coordinates of C are (5, 6).
Hence, $\left(b\right)⇒\left(p\right)$

(c) The points lie on the circle with centre O(2, 3).
Then, we have:
${\left(x-2\right)}^{2}+{\left(5-3\right)}^{2}={\left(4-2\right)}^{2}+{\left(3-3\right)}^{2}$
$⇒{\left(x-2\right)}^{2}+{2}^{2}={2}^{2}+{0}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}=\left({2}^{2}-{2}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒x-2=0\phantom{\rule{0ex}{0ex}}⇒x=2$
Therefore, the value of x = 2
Hence, $\left(c\right)⇒\left(s\right)$

(d) D is the midpoint of BC.
So, the coordinates of D are

Now, the length of the median

Hence, $\left(d\right)⇒\left(q\right)$

#### Question 41:

Assertion (A)
If the points A(8, 1), B(k, −4) and C(2, −5) are collinear, then k = 4.

Reason (R)
Three points A(x1, y1), B(x2, y2) and C(x3, y3) are collinear only when x1(y2 y3) + x2(y3y1) + x3(y1y2) = 0.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

The given points are
Here,
Points A,B and C are collinear.
$⇒{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒8\left(-4+5\right)+k\left(-5-1\right)+2\left(1+4\right)=0\phantom{\rule{0ex}{0ex}}⇒8\left(1\right)+k\left(-6\right)+2\left(5\right)=0\phantom{\rule{0ex}{0ex}}⇒-6k=-18\phantom{\rule{0ex}{0ex}}⇒k=3$
Hence, Assertion(A) is false.
Reason(R):
The given statement is true.
Hence, the correct answer is (d).

#### Question 42:

Assertion (A)
The point P(−1, 6) divides the join of A(−3, 10) and B(6, −8) in the ratio 2 : 7.

Reason (R)
If the point C(x, y) divides the join of A(x1, y1) and B(x2, y2) in the ratio m : n, then

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

The correct answer is (a)
Let the required ratio be k : 1.
Then, by section formula, the coordinates of C are
$P\left(\frac{6k-3}{k+1},\frac{-8k+10}{k+1}\right)$
Therefore,

Therefore, the required ratio is , which is same as 2 : 7.
Hence, Assertion(A) is true.
Reason(R):
The given statement is true.
Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

#### Question 1:

The ratio in which the point P(6, −6) divides the line segment joining A(1, 4) and B(9, −12) is

(a) 2 : 3
(b) 3 : 2
(c) 3 : 5
(d) 5 : 3

(d) 5 : 3
Let the required ratio be k : 1.
Then, by section formula, the coordinates of C are
$P\left(\frac{9k+1}{k+1},\frac{-12k+4}{k+1}\right)\phantom{\rule{0ex}{0ex}}$
Therefore,

Therefore, the required ratio is , which is same as 5 : 3.

#### Question 2:

In what ratio is the segment joining the points A(4, 6) and B(−7, −1) divided by x-axis?

(a) 3 : 1
(b) 6 : 1
(c) 1 : 2
(d) 2 : 3

(b) 6 : 1
Let AB be divided by x-axis in the ratio k : 1 at the point P.
Then, by section formula, the coordinates of P are

But P lies on the x-axis. So, its ordinate is 0.
Therefore,
$\frac{-k+6}{k+1}=0\phantom{\rule{0ex}{0ex}}⇒-k+6=0\phantom{\rule{0ex}{0ex}}⇒k=6$
Hence, the required ratio is 6 : 1.

#### Question 3:

The two vertices of a triangle are A(6, 3) and B(−1, 7) and its centroid is G(1, 5). The third vertex C of ∆ABC is

(a) (2, 5)
(b) (2, −5)
(c) (−2, 5)
(d) (−2, −5)

(c) (−2, 5)
The two vertices of .
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are

But it is given that the centroid is G(1,5).
Therefore,

Hence, the third vertex of

#### Question 4:

The area of ∆ABC with vertices A(1, −1), B(−4, 6) and C(−3, −5) is

(a) 52 sq units
(b) 27 sq units
(c) 48 sq units
(d) 24 sq units

(d) 24 sq units
The given points are .
Here,
Therefore,

#### Question 5:

Find the length of the line segment AB whose end points are A(2, −5) and B(−4, 7).

The given points are .
Then,
Therefore,

#### Question 6:

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

Let the point P(x,0) be equidistant from the points .
Then,
$PA=PB\phantom{\rule{0ex}{0ex}}⇒P{A}^{2}=P{B}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x+3\right)}^{2}+{\left(0-2\right)}^{2}={\left(x-5\right)}^{2}+{\left(0+2\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+6x+13={x}^{2}-10x+29\phantom{\rule{0ex}{0ex}}⇒16x=16\phantom{\rule{0ex}{0ex}}⇒x=1$
Therefore,

#### Question 7:

Find the coordinates of point P that divides the join of A(4, −3) and B(9, 7) in the ratio 3 : 2.

The given points are .
Then,
Also, m = 3 and n = 2
Let the required point be P(x, y).
By section formula, we have:

Hence, the coordinates of P are (7, 3).

#### Question 8:

If A(−2, 4), B(0, 0) and C(4, 2) are the vertices of a ∆ABC, find the length of the median through A.

Let D be the midpoint of BC.
So, the coordinates of D are

Length of the median:

#### Question 9:

If the points A(2, 3), B(4, k) and C(6, −3) are collinear, find the value of k.

The given points are .
Here,
Points A,B and C are collinear. Then,
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒2\left(k+3\right)+4\left(-3-3\right)+6\left(3-k\right)=0\phantom{\rule{0ex}{0ex}}⇒2k+6-24+18-6k=0\phantom{\rule{0ex}{0ex}}⇒-4k=0\phantom{\rule{0ex}{0ex}}⇒k=0$

#### Question 10:

If the points A(4, 3) and B(x, 5) lie on the circle with centre O(2, 3), find the value of x.

The given points are A(4, 3) and B(x, 5), which lie on the circle with centre O(2, 3).
Then, we have:
OB = OA
$⇒\sqrt{{\left(x-2\right)}^{2}+{\left(5-3\right)}^{2}}=\sqrt{{\left(4-2\right)}^{2}+{\left(3-3\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}+{2}^{2}={2}^{2}+{0}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}=\left({2}^{2}-{2}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒x-2=0\phantom{\rule{0ex}{0ex}}⇒x=2$

#### Question 11:

In what ratio does the line xy − 2 = 0 divide the line segment joining the points A(3, −1) and B(8, 9)?

Let the line x − y 2=0 divide the line segment joining A(3, 1) and B(8, 9) in the ratio k : 1 at the point P.
Then, by the section formula, the coordinates of P are
$P\left(\frac{8k+3}{k+1},\frac{9k-1}{k+1}\right)$
Since, P lies on the line x − y 2=0, we have:
$\frac{8k+3}{k+1}-\frac{9k-1}{k+1}-2=0\phantom{\rule{0ex}{0ex}}⇒\left(8k+3\right)-\left(9k-1\right)-\left(2k+2\right)=0\phantom{\rule{0ex}{0ex}}⇒3k=2\phantom{\rule{0ex}{0ex}}⇒k=\frac{2}{3}$
So, the required ratio is ,which is same as 2 : 3.

#### Question 12:

Find the area of ∆ABC with vertices A(1, −3), B(4, −3) and C(−9, 7).

The given points are .
Here,
Therefore,

#### Question 13:

Show that A(−3, −1), B(−4, −1), C(3, 3) and D(4, 3) are the vertices of a rhombus.

Disclaimer:- It should be parallelogram in place of rhombus in the question.
The given vertices are A(−3, −1), B(−4, −1), C(3, 3) and D​(4, 3).
Then,

Here, AB = CD
and  BC = DA
Therefore, ABCD is a parallelogram.

#### Question 14:

Find the points on y-axis, which are at a distance of 13 units each from the point (−5, 7).

Let the point on y-axis be B(0, b) and the given point be A(−5, 7).The distance between A and B is 13 units.
Now,

Hence, the points are (0, 19) and (0, −5).

#### Question 15:

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

Let the fourth vertex be D(a, b).
Join AC and BD.
Let AC and BD intersect at the point O.
We know that the diagonals of a parallelogram bisect each other.

Therefore,

Hence, the fourth vertex is D(3, 2).

#### Question 16:

If the points A(x, y), B(−5, 7) and C(−4, 5) are collinear, show that 2x + y + 3 = 0.

The given points are .
Here,
A,B and C​ are collinear. Then,
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒x\left(7-5\right)-5\left(5-y\right)-4\left(y-7\right)=0\phantom{\rule{0ex}{0ex}}⇒7x-5x-25+5y-4y+28=0\phantom{\rule{0ex}{0ex}}⇒2x+y+3=0$

#### Question 17:

Find the area of quadrilateral ABCD whose vertices are A(−4, −2), B(−3, −5), C(3, −2) and D(2, 3).

Join and C . Then,

Area of quadilateral ABCD

Therefore, area of quadilateral ABCD $=\left(\frac{21}{2}+\frac{35}{2}\right)$

#### Question 18:

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are A(2, 2), B(4, 4) and C(2, 6).

Let A(2, 2), B(4, 4) and C(2, 6) be the vertices of a triangle ABC.
Let D,E and F be the midpoints of AB, BC and CA, respectively.
Then, the coordinates of D, E and F​ are:

#### Question 19:

Prove that the points A(−3, 0), B(1, −3) and C(4, 1) are the vertices of an isosceles right-angled triangle. Find the area of this triangle.

Let  be the given vertices. Then,

Thus, AB = BC = 5 units.
Therefore, $∆ABC$ is an isosceles triangle.
Also,
$\left(A{B}^{2}+B{C}^{2}\right)=\left({5}^{2}+{5}^{2}\right)$
= 50 units.

= 50 units
Thus, $A{B}^{2}+B{C}^{2}=A{{C}^{2}}^{}$
This shows that $∆ABC$ is right-angled at B.
So, ∆ABC is an isosceles right-angled triangle.
In $∆ABC$, we have:
base BC = 5 units and height AB= 5 units

#### Question 20:

If the point P(−1, 2) divides the line segment joining A(2, 5) and B in the ratio 3 : 4, find the coordinates of B.

Let the coordinates of B be .
Then, P divides the line joining A(2, 5) and B in the ratio 3 : 4.
So, by section formula, the coordinates of P are

But, the coordinates of P are (−1, 2).

$⇒3x$1+8=7  and  3y1+20=14

Hence, the coordinates of B are (−5, −2).

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