Rs Aggarwal 2018 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 2018 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 2018 Solutions. All Rs Aggarwal 2018 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 all possible values of y for which the distance between the points is 10 units.

The given points are .

Hence, the possible values of y are .

#### Question 5:

Find the values of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units.

The given points are P(x, 4) and Q(9, 10).

Hence, the values of x are 1 and 17.

#### Question 6:

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

As per the question
$AB=AC\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(x-8\right)}^{2}+{\left(2+2\right)}^{2}}=\sqrt{{\left(x-2\right)}^{2}+{\left(2+2\right)}^{2}}$
Squaring both sides, we get
${\left(x-8\right)}^{2}+{4}^{2}={\left(x-2\right)}^{2}+{4}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-16x+64+16={x}^{2}+4-4x+16\phantom{\rule{0ex}{0ex}}⇒16x-4x=64-4\phantom{\rule{0ex}{0ex}}⇒x=\frac{60}{12}=5$
Now,

Hence, x = 5 and AB = 5 units.

#### Question 7:

If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find the value of p. Also, find the length of AB.

As per the question
$AB=AC\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(0-3\right)}^{2}+{\left(2-p\right)}^{2}}=\sqrt{{\left(0-p\right)}^{2}+{\left(2-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(-3\right)}^{2}+{\left(2-p\right)}^{2}}=\sqrt{{\left(-p\right)}^{2}+{\left(-3\right)}^{2}}$
Squaring both sides, we get
${\left(-3\right)}^{2}+{\left(2-p\right)}^{2}={\left(-p\right)}^{2}+{\left(-3\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒9+4+{p}^{2}-4p={p}^{2}+9\phantom{\rule{0ex}{0ex}}⇒4p=4\phantom{\rule{0ex}{0ex}}⇒p=1$
Now,

Hence, p = 1 and AB$\sqrt{10}$ units.

#### Question 8:

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

Let (x, 0) be the point on the x-axis. Then as per the question, we have

$⇒8x=25-81\phantom{\rule{0ex}{0ex}}⇒x=-\frac{56}{8}=-7$
Hence, the point on the x-axis is (− 7, 0).

#### Question 9:

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

Let P (x, 0) be the point on the x-axis. Then as per the question, we have

Hence, the points on the x-axis are (5, 0) and (17, 0).

#### Question 10:

Find the point on the y-axis which is equidistant from the points A(6, 5) and B(− 4, 3).

Let P (0, y) be a point on the y-axis. Then as per the question, we have

$⇒36+{y}^{2}-10y+25=16+{y}^{2}-6y+9\phantom{\rule{0ex}{0ex}}⇒4y=36\phantom{\rule{0ex}{0ex}}⇒y=9$
Hence, the point on the y-axis is (0, 9).

#### Question 11:

If the point P(x, y) is equidistant from the points A(5, 1) and B(− 1, 5), prove that 3x = 2y.

As per the question, we have

$⇒-10x-2y=2x-10y\phantom{\rule{0ex}{0ex}}⇒8y=12x\phantom{\rule{0ex}{0ex}}⇒3x=2y$
Hence, 3x = 2y.

#### Question 12:

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 13:

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 14:

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 15:

If the point C(− 2, 3) is equidistant from the points A(3, − 1) and B(x, 8), find the value of x. Also, find the distance BC.

As per the question, we have

Now

Hence, x = 2 or −6 and .

#### Question 16:

If the point P(2, 2) is equidistant from the points A(−2, k) and B(−2k, −3), find k. Also find the length of AP.

As per the question, we have

Now for $k=-1$

For $k=-3$

Hence, .

#### Question 17:

If the point (x, y) is equidistant from the points (a + b, ba) and (a b, a + b), prove that bx = ay.

As per the question, we have

$⇒-xa-xb-ay+by=-xa+bx-ya-by\phantom{\rule{0ex}{0ex}}⇒by=bx$
Hence, bx = ay.

#### Question 18:

Using the distance formula, show that the given points are collinear:
(i) (1, −1), (5, 2) and (9, 5)                                         (ii) (6, 9), (0, 1) and (−6, −7)
(iii) (−1, −1), (2, 3) and (8, 11)                                  (iv) (−2, 5), (0, 1) and (2, −3)

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

Hence, the given points are collinear.

(ii)
Let A(6, 9), B(0, 1) and C(−6, −7) be the given points. Then

Hence, the given points are collinear.

(iii)
Let A(−1, −1), B(2, 3) and C(8, 11) be the given points. Then

Hence, the given points are collinear.

(iv)
Let A(−2, 5), B(0, 1) and C(2, −3) be the given points. Then

Hence, the given points are collinear.

#### Question 19:

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 20:

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

The given points are A(3, 0), B(6, 4) and C(− 1, 3). Now

Therefore, A(3, 0), B(6, 4) and C(− 1, 3) are the vertices of an isosceles right triangle.

#### Question 21:

Show that the points A(5, 2), B(2, − 2) and C(− 2, t) are the vertices of a right triangle with $\angle B={90}^{\circ }$, then find the value of t.

$\because \angle B={90}^{\circ }\phantom{\rule{0ex}{0ex}}\therefore A{C}^{2}=A{B}^{2}+B{C}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(5+2\right)}^{2}+{\left(2-t\right)}^{2}={\left(5-2\right)}^{2}+{\left(2+2\right)}^{2}+{\left(2+2\right)}^{2}+{\left(-2-t\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(7\right)}^{2}+{\left(t-2\right)}^{2}={\left(3\right)}^{2}+{\left(4\right)}^{2}+{\left(4\right)}^{2}+{\left(t+2\right)}^{2}$
$⇒49+{t}^{2}-4t+4=9+16+16+{t}^{2}+4t+4\phantom{\rule{0ex}{0ex}}⇒8-4t=4t\phantom{\rule{0ex}{0ex}}⇒8t=8\phantom{\rule{0ex}{0ex}}⇒t=1$
Hence, t = 1.

#### Question 22:

Prove that the points A(2, 4), B(2, 6) and are the vertices of an equilateral triangle.

The given points are A(2, 4), B(2, 6) and . Now

Hence, the points A(2, 4), B(2, 6) and are the vertices of an equilateral triangle.

#### Question 23:

Show that the points (− 3, − 3), (3, 3) and are the vertices of an equilateral triangle.

Let the given points be A(− 3, − 3), B(3, 3) and C. Now

Hence, the given points are the vertices of an equilateral triangle.

#### Question 24:

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 25:

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 26:

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

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

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

Therefore, the given points form a square.

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

Therefore, the given points form a square.

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

Therefore, the given points form a square.

#### Question 27:

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 28:

Show that the points A(3, 0), B(4, 5), C(−1, 4) and D(−2, −1) are the vertices of a rhombus. Find its area.

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

Therefore, the given points are the vertices of a rhombus.

Hence, the area of the rhombus is 24 sq. units.

#### Question 29:

Show that the points A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of a rhombus. Find its area.

The given points are A(6, 1), B(8, 2), C(9, 4) and D(7, 3).

Therefore, the given points are the vertices of a rhombus. Now

Hence, the area of the rhombus is 3 sq. units.

#### Question 30:

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 31:

Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a parallelogram. Show that ABCD is not a rectangle.

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

Therefore, ABCD is a parallelogram. Now

Thus, the diagonals AC and BD are not equal and hence ABCD is not a rectangle.

#### Question 32:

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

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

(i) 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.

(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(0, −4), B(6, 2) C(3, 5) and D(−3, −1).

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

#### 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:

If the coordinates of points A and B are (−2, −2) and (2, −4) respectively, find the coordinates of the point P
such that $AP=\frac{3}{7}AB$, where P lies on the line segment AB.     [CBSE 2015]

The coordinates of the points A and B are (−2, −2) and (2, −4) respectively, where $AP=\frac{3}{7}AB$ and P lies on the line segment AB. So

Let (x, y) be the coordinates of P which divides AB in the ratio 3 : 4 internally. Then
$x=\frac{3×2+4×\left(-2\right)}{3+4}=\frac{6-8}{7}=-\frac{2}{7}\phantom{\rule{0ex}{0ex}}y=\frac{3×\left(-4\right)+4×\left(-2\right)}{3+4}=\frac{-12-8}{7}=-\frac{20}{7}$
Hence, the coordinates of point P are .

#### Question 4:

Point A lies on the line segment PQ joining P(6, −6) and Q(−4, −1) in such a way that $\frac{PA}{PQ}=\frac{2}{5}$.
If the point A also lies  on the line 3x + k (y + 1) = 0, find the value of k.                               [CBSE 2015]

Let the coordinates of A be (x, y). Here, $\frac{PA}{PQ}=\frac{2}{5}$. So,

Let (x, y) be the coordinates of A, which divides PQ in the ratio 2 : 3 internally. Then using section formula, we get
$x=\frac{2×\left(-4\right)+3×\left(6\right)}{2+3}=\frac{-8+18}{5}=\frac{10}{5}=2\phantom{\rule{0ex}{0ex}}y=\frac{2×\left(-1\right)+3×\left(-6\right)}{2+3}=\frac{-2-18}{5}=\frac{-20}{5}=-4$
Now, the point (2, −4) lies on the line 3x + k (y + 1) = 0, therefore
$3×2+k\left(-4+1\right)=0\phantom{\rule{0ex}{0ex}}⇒3k=6\phantom{\rule{0ex}{0ex}}⇒k=\frac{6}{3}=2$
Hence, k = 2.

#### Question 5:

Point P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in five equal parts.
Find the coordinates of the points P, Q and R.

Since, the points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in five equal parts, so
AP = PQ = QR = RS = SB
Here, point P divides AB in the ratio of 1 : 4 internally. So using section formula, we get

The point Q divides AB in the ratio of 2 : 3 internally. So using section formula, we get

The point R divides AB in the ratio of 3 : 2 internally. So using section formula, we get

Hence, the coordinates of the points P, Q and R are (2, 3), (3, 4) and (4, 5) respectively.

#### Question 6:

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 7:

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 8:

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 9:

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 10:

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 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 divides the line segment joining the points
and $\left(2,-5\right)$.                                        [CBSE 2015]

Let k : 1 be the ratio in which the point divides the line segment joining the points and $\left(2,-5\right)$. Then

Hence, the required ratio is 1 : 5.

#### 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 27:

In what ratio does y-axis divide the line segment joining the points (−4, 7) and (3, −7)?     [CBSE 2012]

Let y-axis divides the line segment joining the points (−4, 7) and (3, −7) in the ratio k : 1. Then
$0=\frac{3k\mathit{-}4}{k+1}\phantom{\rule{0ex}{0ex}}⇒3k=4\phantom{\rule{0ex}{0ex}}⇒k=\frac{4}{3}$
Hence, the required ratio is 4 : 3.

#### Question 28:

If the point lies on the line segment joining the points A(3, −5) and B(−7, 9) then find the
ratio in which P divides AB. Also, find the value of y.

Let the point  divides the line segment joining the points A(3, −5) and B(−7, 9) in the ratio k : 1. Then

Now, substituting $k=\frac{1}{3}$ in $\frac{9k-5}{k+1}=y$, we get
$\frac{\frac{9}{3}-5}{\frac{1}{3}+1}=y⇒y=\frac{9-15}{1+3}=-\frac{3}{2}$
Hence, required ratio is 1 : 3 and $y=-\frac{3}{2}$.

#### Question 29:

Find the ratio in which the line segment joining the points A(3, −3) and B(−2, 7) is divided by x-axis.
Also, find the point of division.                                [CBSE 2015]

The line segment joining the points A(3, − 3) and B(− 2, 7) is divided by x-axis. Let the required ratio be k : 1. So,
$0=\frac{k\left(7\right)-3}{k+1}⇒k=\frac{3}{7}$
Now

Hence, the required ratio is 3 : 7 and the point of division is .

#### Question 30:

The base QR of an equilateral triangle PQR lies on x-axis. The coordinates of the point Q are (−4, 0)
and origin is the midpoint of the base. Find the coordinates of the points P and R.

Let (x, 0) be the coordinates of R. Then
$0=\frac{-4+x}{2}⇒x=4$
Thus, the coordinates of R are (4, 0).
Here, PQ = QR = PR and the coordinates of P lies on y-axis. Let the coordinates of P be (0, y). Then
$PQ=QR⇒P{Q}^{2}=Q{R}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(0+4\right)}^{2}+{\left(y-0\right)}^{2}={8}^{2}\phantom{\rule{0ex}{0ex}}⇒{y}^{2}=64-16=48\phantom{\rule{0ex}{0ex}}⇒y=±4\sqrt{3}$
Hence, the required coordinates are .

#### Question 31:

The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, −3).
The origin is the midpoint of the base. Find the coordinates of the points A and B. Also, find
the coordinates of another point D such that ABCD is a rhombus.

Let (0, y) be the coordinates of B. Then
$0=\frac{-3+y}{2}⇒y=3$
Thus, the coordinates of B are (0, 3).
Here, AB = BC = AC and by symmetry the coordinates of A lies on x-axis. Let the coordinates of A be (x, 0). Then
$AB=BC⇒A{B}^{2}=B{C}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(0-3\right)}^{2}={6}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=36-9=27\phantom{\rule{0ex}{0ex}}⇒x=±3\sqrt{3}$
If the coordinates of point A are , then the coordinates of D are .
If the coordinates of point A are , then the coordinates of D are .
Hence, the required coordinates are or .

#### Question 32:

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

Let k be the ratio in which P( −1, y) divides the line segment joining the points A(−3, 10) and B(6, −8). Then

Substituting $k=\frac{2}{7}$ in , we get
$y=\frac{\frac{-8×2}{7}+10}{\frac{2}{7}+1}=\frac{-16+70}{9}=6$
Hence, the required ratio is 2 : 7 and y = 6.

#### Question 33:

ABCD is a rectangle formed by the points . If P, Q, R and S be the mid points of
AB, BC, CD and DA respectively, show that PQRS is a rhombus.

Here, the points P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Then

Now
$PQ=\sqrt{{\left(2+1\right)}^{2}+{\left(4-\frac{3}{2}\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\phantom{\rule{0ex}{0ex}}QR=\sqrt{{\left(5-2\right)}^{2}+{\left(\frac{3}{2}-4\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\phantom{\rule{0ex}{0ex}}RS=\sqrt{{\left(5-2\right)}^{2}+{\left(\frac{3}{2}+1\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\phantom{\rule{0ex}{0ex}}SP=\sqrt{{\left(2+1\right)}^{2}+{\left(-1-\frac{3}{2}\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\phantom{\rule{0ex}{0ex}}PR=\sqrt{{\left(5+1\right)}^{2}+{\left(\frac{3}{2}-\frac{3}{2}\right)}^{2}}=\sqrt{36}=6\phantom{\rule{0ex}{0ex}}QS=\sqrt{{\left(2-2\right)}^{2}+{\left(-1-4\right)}^{2}}=\sqrt{25}=5\phantom{\rule{0ex}{0ex}}$
Thus, PQ = QR = RS = SP and $PR\ne QS$ therefore PQRS is a rhombus.

#### Question 34:

The midpoint P of the line segment joining the points A(−10, 4) and B(−2, 0) lies on the line segment joining the points C(−9, −4) and D(−4, y). Find the ratio in which P divides CD. Also find the value of y.

The midpoint of AB is .
Let k be the ratio in which P divides CD. So

Now, substituting $k=\frac{3}{2}$ in $\frac{k\left(\mathrm{y}\right)-4}{k+1}=2$, we get
$\frac{y×\frac{3}{2}-4}{\frac{3}{2}+1}=2\phantom{\rule{0ex}{0ex}}⇒\frac{3y-8}{5}=2\phantom{\rule{0ex}{0ex}}⇒y=\frac{10+8}{3}=6$
Hence, the required ratio is 3 : 2 and y = 6.

#### 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:

Find the area of quadrilateral ABCD whose vertices are A(3, −1), B(9, −5), C(14, 0) and D(9, 19).                [CBSE 2012]

By joining A and C, we get two triangles ABC and ACD.
Let . Then

So, the area of the quadrilateral is 25 + 107 = 132 sq. units.

#### Question 3:

Find the area of quadrilateral PQRS whose vertices are P(−5, −3), Q(−4, −6), R(2, −3) and S(1, 2).
[CBSE 2015]

By joining P and R, we get two triangles PQR and PRS.
Let . Then

So, the area of the quadrilateral PQRS is  sq. units.

#### Question 4:

Find the area of quadrilateral ABCD whose vertices are A(−3, −1), B(−2, −4), C(4, −1) and D(3, 4).
[CBSE 2013C]

By joining A and C, we get two triangles ABC and ACD.
Let . Then

So, the area of the quadrilateral ABCD is  sq. units.

#### Question 5:

Find the area of quadrilateral ABCD whose vertices are A(−5, 7), B(−4, −5), C(−1, −6) and D(4, 5).

By joining A and C, we get two triangles ABC and ACD.
Let . Then

So, the area of the quadrilateral ABCD is .

#### Question 6:

Find area of the triangle formed by joining the midpoints of the sides of the triangle
whose vertices are A(2, 1), B(4, 3) and C(2, 5).

The vertices of the triangle are A(2, 1), B(4, 3) and C(2, 5).

Now

Hence, the area of the required triangle is 1 sq. unit.

#### Question 7:

A(7, −3), B(5, 3), C(3, −1) are the vertices of a $∆ABC$ and AD is its median. Prove that the median
AD divides $∆ABC$ into two triangles of equal areas.

The vertices of the triangle are A(7, −3), B(5, 3), C(3, −1).

For the area of the triangle ADC, let . Then

Now, for the area of triangle ABD, let . Then

Thus, .
Hence, AD divides $∆ABC$ into two triangles of equal areas.

#### Question 8:

Find the area of $∆ABC$ with A(1, −4) and midpoints of sides through A being (2, −1) and (0, −1).      [CBSE 2015]

Let be the coordinates of B and C respectively. Since, the coordinates of A are (1, −4), therefore
$\frac{1+{x}_{2}}{2}=2⇒{x}_{2}=3\phantom{\rule{0ex}{0ex}}\frac{-4+{y}_{2}}{2}=-1⇒{y}_{2}=2\phantom{\rule{0ex}{0ex}}\frac{1+{x}_{3}}{2}=0⇒{x}_{3}=-1\phantom{\rule{0ex}{0ex}}\frac{-4+{y}_{3}}{2}=-1⇒{y}_{3}=2$
Let . Now

Hence, the area of the triangle $∆ABC$ is 12 sq. units.

#### Question 9:

A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of $∆ADE$.                                                                                                                                           [CBSE 2015]

Let (x, y) be the coordinates of D and be the coordinates of E. Since, the diagonals of a parallelogram bisect
each other at the same point, therefore
$\frac{x+8}{2}=\frac{6+9}{2}⇒x=7\phantom{\rule{0ex}{0ex}}\frac{y+2}{2}=\frac{1+4}{2}⇒y=3$
Thus, the coordinates of D are (7, 3).
E is the midpoint of DC, therefore
$x\text{'}=\frac{7+9}{2}⇒x\text{'}=8\phantom{\rule{0ex}{0ex}}y\text{'}=\frac{3+4}{2}⇒y\text{'}=\frac{7}{2}$
Thus, the coordinates of E are .
Let . Now

Hence, the area of the triangle $∆ADE$ is .

#### Question 10:

If the vertices of $∆ABC$ be A(1, −3), B(4, p) and C(−9, 7) and its area is 15 square units, find the values of p.

Let . Now
$\mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{\mathit{1}}\mathit{-}{y}_{\mathit{2}}\right)\right]\phantom{\rule{0ex}{0ex}}⇒15=\frac{1}{2}\left[1\left(p-7\right)+4\left(7+3\right)-9\left(-3-p\right)\right]\phantom{\rule{0ex}{0ex}}⇒15=\frac{1}{2}\left[10p+60\right]\phantom{\rule{0ex}{0ex}}⇒\left|10p+60\right|=30$
Therefore

Hence, .

#### Question 11:

Find the value of k so that the area of the triangle with vertices A(k + 1, 1), B(4, −3) and C(7, −k) is 6 square units.

Let . Now
$\mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}\mathit{-}{y}_{2}\right)\right]\phantom{\rule{0ex}{0ex}}⇒6=\frac{1}{2}\left[\left(k+1\right)\left(-3+k\right)+4\left(-k-1\right)+7\left(1+3\right)\right]\phantom{\rule{0ex}{0ex}}⇒6=\frac{1}{2}\left[{k}^{2}-2k-3-4k-4+28\right]\phantom{\rule{0ex}{0ex}}⇒{k}^{2}-6k+9=0$
$⇒{\left(k-3\right)}^{2}=0⇒k=3$
Hence, k = 3.

#### Question 12:

For what value of k (k > 0) is the area of the triangle with vertices (−2, 5), (k, −4) and
(2k + 1, 10) equal to 53 square units?

Let be the vertices of
the triangle. So
$\mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]\phantom{\rule{0ex}{0ex}}⇒53=\frac{1}{2}\left[\left(-2\right)\left(-4-10\right)+k\left(10-5\right)+\left(2k+1\right)\left(5+4\right)\right]\phantom{\rule{0ex}{0ex}}⇒53=\frac{1}{2}\left[28+5k+9\left(2k+1\right)\right]\phantom{\rule{0ex}{0ex}}⇒28+5k+18k+9=106$

$⇒37+23k=106\phantom{\rule{0ex}{0ex}}⇒23k=106-37=69\phantom{\rule{0ex}{0ex}}⇒k=\frac{69}{23}=3$
Hence, k = 3.

#### Question 13:

Show that the following points are collinear:

(i) A(2, −2), B(−3, 8) and C(−1, 4)
(ii) A(−5, 1), B(5, 5) and C(10, 7)
(iii) A(5, 1), B(1, −1) and C(11, 4)
(iv) A(8, 1), B(3, −4) and C(2, −5)

(i)
Let A(x1 = 2, y1 = −2), B(x2 = −3, y2 = 8) and C(x3 = −1, y3 = 4) be the given points. Now
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=2\left(8-4\right)+\left(-3\right)\left(4+2\right)+\left(-1\right)\left(-2-8\right)\phantom{\rule{0ex}{0ex}}=8-18+10\phantom{\rule{0ex}{0ex}}=0$
Hence, the given points are collinear.

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

(iii)
Let A(x1 = 5, y1 = 1), B(x2 = 1, y2 = −1) and C(x3 = 11, y3 = 4) be the given points. Now
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=5\left(-1-4\right)+1\left(4-1\right)+11\left(1+1\right)\phantom{\rule{0ex}{0ex}}=-25+3+22\phantom{\rule{0ex}{0ex}}=0$
Hence, the given points are collinear.

(iv)
Let A(x1 = 8, y1 = 1), B(x2 = 3, y2 = −4) and C(x3 = 2, y3 = −5) be the given points. Now
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=8\left(-4+5\right)+3\left(-5-1\right)+2\left(1+4\right)\phantom{\rule{0ex}{0ex}}=8-18+10\phantom{\rule{0ex}{0ex}}=0$
Hence, the given points are collinear.

#### Question 14:

Find the value of x for which the points are collinear.      [CBSE 2015]

Let . So, the condition for three collinear points is
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}\mathit{-}{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒x\left(-4+5\right)-3\left(-5-2\right)+7\left(2+4\right)=0\phantom{\rule{0ex}{0ex}}⇒x+21+42=0\phantom{\rule{0ex}{0ex}}⇒x=-63$
Hence, x = − 63.

#### Question 15:

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 16:

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 17:

Find the value of y for which the points A(−3, 9), B(2, y) and C(4, −5) are collinear.

Let A(x1 = −3, y1 = 9), B(x2 = 2, y2 = y) and C(x3 = 4, y3 = −5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(-3\right)\left(y+5\right)+2\left(-5-9\right)+4\left(9-y\right)=0\phantom{\rule{0ex}{0ex}}⇒-3y-15-28+36-4y=0\phantom{\rule{0ex}{0ex}}⇒7y=36-43$
$⇒y=-1$

#### Question 18:

For what values of k are the points A(8, 1), B(3, −2k) and C(k, −5) collinear?                  [CBSE 2015]

Let A(x1 = 8, y1 = 1), B(x2 = 3, y2 = −2k) and C(x3 = ky3 = −5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒8\left(-2k+5\right)+3\left(-5-1\right)+k\left(1+2k\right)=0\phantom{\rule{0ex}{0ex}}⇒-16k+40-18+k+2{k}^{2}=0\phantom{\rule{0ex}{0ex}}⇒2{k}^{2}-15k+22=0$

Hence, .

#### Question 19:

Find a relation between x and y, if the points A(2, 1), B(x, y) and C(7, 5) are collinear.                 [CBSE 2009C]

Let A(x1 = 2, y1 = 1), B(x2 = xy2 = y) and C(x3 = 7, y3 = 5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒2\left(y-5\right)+x\left(5-1\right)+7\left(1-y\right)=0\phantom{\rule{0ex}{0ex}}⇒2y-10+4x+7-7y=0\phantom{\rule{0ex}{0ex}}⇒4x-5y-3=0$
Hence, the required relation is 4x − 5y − 3 = 0.

#### Question 20:

Find a relation between x and y, if the points A(x, y), B(5, 7) and C(4, 5) are collinear.      [CBSE 2015]

Let A(x1 = xy1 = y), B(x2 = −5, y2 = 7) and C(x3 = −4, y3 = 5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒x\left(7-5\right)+\left(-5\right)\left(5-y\right)+\left(-4\right)\left(y-7\right)=0\phantom{\rule{0ex}{0ex}}⇒7x-5x-25+5y-4y+28=0\phantom{\rule{0ex}{0ex}}⇒2x+y+3=0$
Hence, the required relation is 2x + y + 3 = 0.

#### Question 21:

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 22:

If the points P(−3, 9), Q(a, b) and R(4, 5) are collinear and a + b = 1, find the values of a and b.     [CBSE 2014]

Let A(x1 = −3, y1 = 9), B(x2 = ay2 = b) and C(x3 = 4, y3 = 5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(-3\right)\left(b+5\right)+a\left(-5-9\right)+4\left(9-b\right)=0\phantom{\rule{0ex}{0ex}}⇒-3b-15-14a+36-4b=0\phantom{\rule{0ex}{0ex}}⇒2a+b=3$
Now, solving a + b = 1 and 2a + b = 3, we get a = 2 and b = −1.
Hence, a = 2 and b = −1.

#### Question 23:

Find the area of $∆ABC$ with vertices A(0, −1), B(2, 1) and C(0, 3). Also, find the area of the triangle formed
by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is 4 : 1.           [CBSE 2014]

Let A(x1 = 0, y1 = −1), B(x2 = 2, y2 = 1) and C(x3 = 0, y3 = 3) be the given points. Then

So, the area of the triangle $∆ABC$ is 4 sq. units.
Let D(a1, b1), E(a2, b2) and F(a3, b3) be the midpoints of AB, BC and AC respectively. Then

Thus, the coordinates of D, E and F are D(a1 = 1, b1 = 0), E(a2 = 1, b2 = 2) and F(a3 = 0, b3 = 1). Now

So, the area of the triangle $∆DEF$ is 1 sq. unit.
Hence, .

#### Question 1:

Points A(−1, y) and B(5, 7) lie on a circle with centre O(2, −3y). Find the values of y.     [CBSE 2014]

The given points are A(−1, y), B(5, 7) and O(2, −3y).
Here, AO and BO are the radii of the circle. So
$AO=BO⇒A{O}^{2}=B{O}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(2+1\right)}^{2}+{\left(-3y-y\right)}^{2}={\left(2-5\right)}^{2}+{\left(-3y-7\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒9+{\left(4y\right)}^{2}={\left(-3\right)}^{2}+{\left(3y+7\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒9+16{y}^{2}=9+9{y}^{2}+49+42y$
$⇒7{y}^{2}-42y-49=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-6y-7=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-7y+y-7=0\phantom{\rule{0ex}{0ex}}⇒y\left(y-7\right)+1\left(y-7\right)=0$

Hence, y = 7 or y = −1.

#### Question 2:

If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p.                [CBSE 2014]

The given points are A(0, 2), B(3, p) and C(p, 5).
$AB=AC⇒A{B}^{2}=A{C}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(3-0\right)}^{2}+{\left(p-2\right)}^{2}={\left(p-0\right)}^{2}+{\left(5-2\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒9+{p}^{2}-4p+4={p}^{2}+9\phantom{\rule{0ex}{0ex}}⇒4p=4⇒p=1$
Hence, p = 1.

#### Question 3:

ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). Find the length of one of its diagonal.
[CBSE 2014]

The given vertices are B(4, 0), C(4, 3) and D(0, 3). Here, BD is one of the diagonals. So

Hence, the length of the diagonal is 5 units..

#### Question 4:

If the point P(k − 1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k.             [CBSE 2014]

The given points are P(k − 1, 2), A(3, k) and B(k, 5).

Hence, k = 1 or k = 5.

#### Question 5:

Find the ratio in which the point P(x, 2) divides the join of A(12, 5) and B(4, −3).          [CBSE 2014]

Let k be the ratio in which the point P(x, 2) divides the line joining the points A(x1 = 12, y1 = 5) and B(x2 = 4, y2 = −3). Then

Now
$2=\frac{k×\left(-3\right)+5}{k+1}⇒2k+2=-3k+5⇒k=\frac{3}{5}\phantom{\rule{0ex}{0ex}}$
Hence, the required ratio is 3 : 5.

#### Question 6:

Prove that the diagonals of a rectangle ABCD with vertices A(2, −1), B(5, −1), C(5, 6) and D(2, 6) are
equal and bisect each other.                                         [CBSE 2014]

The vertices of the rectangle ABCD are A(2, −1), B(5, −1), C(5, 6) and D(2, 6). Now

Since, the midpoints of AC and BD coincide, therefore the diagonals of rectangle ABCD bisect each other.

#### Question 7:

Find the lengths of the medians AD and BE of $∆ABC$ whose vertices are A(7, −3), B(5, 3) and C(3, −1).       [CBSE 2014]

The given vertices are A(7, −3), B(5, 3) and C(3, −1).
Since D and E are the midpoints of BC and AC respectively, therefore

Now
$AD=\sqrt{{\left(7-4\right)}^{2}+{\left(-3-1\right)}^{2}}=\sqrt{9+16}=5\phantom{\rule{0ex}{0ex}}BE=\sqrt{{\left(5-5\right)}^{2}+{\left(3+2\right)}^{2}}=\sqrt{0+25}=5$
Hence, AD = BE = 5 units.

#### Question 8:

If the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3 then find the value of k.   [CBSE 2013C]

Here, the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3. So

Hence, $k=\frac{16}{5}$.

#### Question 9:

Find the point on x-axis which is equidistant from points A(−1, 0) and B(5, 0).               [CBSE 2013C]

Let P(x, 0) be the point on x-axis. Then
$AP=BP⇒A{P}^{2}=B{P}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x+1\right)}^{2}+{\left(0-0\right)}^{2}={\left(x-5\right)}^{2}+{\left(0-0\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x+1={x}^{2}-10x+25\phantom{\rule{0ex}{0ex}}⇒12x=24⇒x=2$
Hence, x = 2.

#### Question 10:

Find the distance between the points

The given points are .
Then,
Therefore,

#### Question 11:

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 12:

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 13:

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 14:

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 15:

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 16:

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 17:

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:

The distance of the point P(−6, 8) from the origin is                              [CBSE 2013C]
(a) 8                       (b) $2\sqrt{7}$                         (c) 6                               (d) 10

The distance of a point (x, y) from the origin O(0, 0) is $\sqrt{{x}^{2}+{y}^{2}}$.
Let P(x = −6, y = 8) be the given point. Then

Hence, the correct answer is option (d).

#### Question 2:

The distance of the point (−3, 4) from x-axis is                            [CBSE 2012]
(a) 3                              (b) −3                                 (c) 4                                      (d) 5

The distance of a point (x, y) from x-axis is $\left|y\right|$.
Here, the point is (−3, 4). So, its distance from x-axis is $\left|4\right|=4$.
Hence, the correct answer is option (c).

#### Question 3:

The point on x-axis which is equidistant from points A(−1, 0) and B(5, 0) is                  [CBSE 2013]
(a) (0, 2)                                (b) (2, 0)                                      (c) (3, 0)                                              (d) (0, 3)

Let P(x, 0) the point on x-axis, then
$AP=BP⇒A{P}^{2}=B{P}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x+1\right)}^{2}+{\left(0-0\right)}^{2}={\left(x-5\right)}^{2}+{\left(0-0\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x+1={x}^{2}-10x+25\phantom{\rule{0ex}{0ex}}⇒12x=24⇒x=2$
Thus, the required point is (2, 0).
Hence, the correct answer is option (b).

#### Question 4:

If R(5, 6) is the midpoint of the line segment AB joining the points A(6, 5) and B(4, y) then y equals
(a) 5                             (b) 7                                (c) 12                                (d) 6

Since R(5, 6) is the midpoint of the line segment AB joining the points A(6, 5) and B(4, y), therefore
$\frac{5+y}{2}=6\phantom{\rule{0ex}{0ex}}⇒5+y=12\phantom{\rule{0ex}{0ex}}⇒y=12-5=7$
Hence, the correct option is (b).

#### Question 5:

If the point C(k, 4) divides the join of the points A(2, 6) and B(5, 1) in the ratio 2 : 3 then the value of k is                                                                                                                                                [CBSE 2013C]
(a) 16                                 (b) $\frac{28}{5}$                                     (c) $\frac{16}{5}$                                    (d) $\frac{8}{5}$

The point C(k, 4) divides the join of the points A(2, 6) and B(5, 1) in the ratio 2 : 3. So
$k=\frac{2×5+3×2}{2+3}=\frac{10+6}{5}=\frac{16}{5}$
Hence, the correct answer is option (c).

#### Question 6:

The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is                        [CBSE 2014]
(a) $7+\sqrt{5}$                                 (b) 5                                     (c) 10                                    (d) 12

Let A(0, 4), B(0, 0) and C(3, 0) be the given vertices. So
$AB=\sqrt{{\left(0-0\right)}^{2}+{\left(4-0\right)}^{2}}=\sqrt{16}=4\phantom{\rule{0ex}{0ex}}BC=\sqrt{{\left(0-3\right)}^{2}+{\left(0-0\right)}^{2}}=\sqrt{9}=3\phantom{\rule{0ex}{0ex}}AC=\sqrt{{\left(0-3\right)}^{2}+{\left(4-0\right)}^{2}}=\sqrt{9+16}=5$
Therefore
AB + BC + AC = 4 + 3 + 5 = 12
Hence, the correct answer is option (d).

#### Question 7:

If A(1, 3), B(−1, 2) and C(2, 5) and D(x, 4) are the vertices of a ||gm ABCD then the value of x is       [CBSE 2012]
(a) 3                                 (b) 4                                     (c) 0                                    (d) $\frac{3}{2}$

The diagonals of a parallelogram bisect each other. The vertices of the ||gm ABCD are A(1, 3), B(−1, 2) and C(2, 5) and D(x, 4).
Here, AC and BD are the diagonals. So
$\frac{1+2}{2}=\frac{-1+x}{2}\phantom{\rule{0ex}{0ex}}⇒x-1=3\phantom{\rule{0ex}{0ex}}⇒x=1+3=4$
Hence, the correct answer is option (b).

#### Question 8:

If the points A(x, 2), B(−3, −4) and C(7, −5) are collinear then the value of x is                        [CBSE 2014]
(a) −63                                 (b) 63                                     (c) 60                                    (d) −60

Let A(x1 = x, y1 = 2), B(x2 = −3, y2 = −4) and C(x3 = 7, y3 = −5) be collinear points. Then
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒x\left(-4+5\right)+\left(-3\right)\left(-5-2\right)+7\left(2+4\right)=0\phantom{\rule{0ex}{0ex}}⇒x+21+42=0\phantom{\rule{0ex}{0ex}}⇒x=-63$
Hence, the correct answer is option (a).

#### Question 9:

The area of a triangle with vertices A(5, 0), B(8, 0) and C(8, 4) in square units is                   [CBSE 2012]
(a) 20                                 (b) 12                                     (c) 6                                    (d) 16

Let A(x1 = 5, y1 = 0), B(x2 = 8, y2 = 0) and C(x3 = 8, y3 = 4) be the vertices of the triangle. Then

Hence, the correct answer is option (c).

#### Question 10:

The area of $∆ABO$ with vertices A(a, 0), O(0, 0) and B(0, b) in square units is
(a) ab                                  (b) $\frac{1}{2}ab$                                 (c) $\frac{1}{2}{a}^{2}{b}^{2}$                                     (d) $\frac{1}{2}{b}^{2}$

Let A(x1 = a, y1 = 0), O(x2 = 0, y2 = 0) and B(x3 = 0, y3 = b) be the given vertices. So

Hence, the correct answer is (b).

#### Question 11:

If is the midpoint of the line segment joining the points A(−6, 5) and B(−2, 3) then the value of a is  [CBSE 2011]
(a) −8                                  (b) 3                                 (c) −4                                     (d) 4

The point is the midpoint of the line segment joining the points A(−6, 5) and B(−2, 3). So
$\frac{a}{2}=\frac{-6-2}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{2}=-4\phantom{\rule{0ex}{0ex}}⇒a=-8$
Hence, the correct answer is option (a).

#### Question 12:

ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). The length of one of its diagonals is  [CBSE 2014]
(a) 5                                  (b) 4                                 (c) 3                                     (d) 25

Here, AC and BD are two diagonals of the rectangle ABCD. So

Hence, the correct answer is option (a).

#### Question 13:

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 is    [CBSE 2012]
(a) (2, 4)                                  (b) (3, 5)                                 (c) (4, 2)                                     (d) (5, 3)

Here, the point P divides the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1. Then

Hence, the correct answer is option (b).

#### Question 14:

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                     [CBSE 2012]
(a) (−6, 7)                                  (b) (6, −7)                                 (c) (4, 2)                                     (d) (5, 3)

Let (x, y) be the coordinates of the other end of the diameter. Then
$-2=\frac{2+x}{2}⇒x=-6\phantom{\rule{0ex}{0ex}}5=\frac{3+y}{2}⇒y=7$
Hence, the correct answer is option (a).

#### Question 15:

In the given figure P(5, −3) and Q(3, y) are the points of trisection of the line segment
joining A(7, −2) and B(1, −5). Then, y equals                                 [CBSE 2012]

(a) 2                                  (b) 4                                 (c) −4                                     (d) $\frac{-5}{2}$

Here, AQ : BQ = 2 : 1. Then

Hence, the correct answer is option (c).

#### Question 16:

The midpoint of segment AB is P(0, 4). If the coordinates of B are (−2, 3), then the coordinates of A are                [CBSE 2011]
(a) (2, 5)                             (b) (−2, −5)                               (c) (2, 9)                                  (d) (−2, 11)

Let (x, y) be the coordinates of A. Then
$0=\frac{-2+x}{2}⇒x=2\phantom{\rule{0ex}{0ex}}4=\frac{3+y}{2}⇒y=8-3=5$
Thus, the coordinates of A are (2, 5).
Hence, the correct answer is option (a).

#### Question 17:

The point P which divides the line segment joining the points A(2, −5) and B(5, 2) in the ratio 2 : 3 lies in the quadrant                                                [CBSE 2011]
(a) I                           (b) II                             (c) III                           (d) IV

Let (x, y) be the coordinates of P. Then
$x=\frac{2×5+3×2}{2+3}=\frac{10+6}{5}=\frac{16}{5}\phantom{\rule{0ex}{0ex}}y=\frac{2×2+3×\left(-5\right)}{2+3}=\frac{4-15}{5}=\frac{-11}{5}$
Thus, the coordinates of point P are and so it lies in the fourth quadrant.
Hence, the correct answer is option (d).

#### Question 18:

If A(−6, 7) and B(−1, −5) are two given points then the distance 2AB is                             [CBSE 2011]
(a) 13                           (b) 26                             (c) 169                           (d) 238

The given points are A(−6, 7) and B(−1, −5). So

Thus, 2AB = 26.
Hence, the correct answer is option (b).

#### Question 19:

Which point on 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)

Let P(x, 0) be the point on x-axis. Then as per the question
$AP=BP⇒A{P}^{2}=B{P}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-7\right)}^{2}+{\left(0-6\right)}^{2}={\left(x+3\right)}^{2}+{\left(0-4\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-14x+49+36={x}^{2}+6x+9+16\phantom{\rule{0ex}{0ex}}⇒60=20x\phantom{\rule{0ex}{0ex}}⇒x=\frac{60}{20}=3$
Thus, the required point is (3, 0).
Hence, the correct answer is option (c).

#### Question 20:

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 21:

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 22:

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 23:

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 24:

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 25:

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 26:

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 27:

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 28:

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 29:

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 30:

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 31:

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 32:

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 33:

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 34:

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

$AB=5\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}=5\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(1-4\right)}^{2}+{\left(0-p\right)}^{2}}=5\phantom{\rule{0ex}{0ex}}⇒{\left(-3\right)}^{2}+{\left(-p\right)}^{2}=25\phantom{\rule{0ex}{0ex}}⇒9+{p}^{2}=25\phantom{\rule{0ex}{0ex}}⇒{p}^{2}=16\phantom{\rule{0ex}{0ex}}⇒p=±\sqrt{16}\phantom{\rule{0ex}{0ex}}⇒p=±4\phantom{\rule{0ex}{0ex}}$