CBSE Class 10 Coordinate geometry Sure Shot Questions Set D

Points to Remember

1. The distance between two points, \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is given by
\( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

2. If \( C(x, y) \) divides the line joining \( A(x_1, y_1) \) and \( B(x_2, y_2) \) internally in the ratio \( m_1 : m_2 \), then the coordinates of \( C(x, y) \) are
\( (x, y) = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right) \)

3. The coordinates of mid-point \( C(x, y) \) of a line segment joining the points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) are given by,
\( (x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)

4. The coordinates of centroid of a triangle whose vertices are \( (x_1, y_1), (x_2, y_2) \) and \( (x_3, y_3) \), are given as,
\( (x, y) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)

5. The area of a triangle whose vertices are \( (x_1, y_1), (x_2, y_2) \) and \( (x_3, y_3) \), is given by
\( \Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)

Multiple Choice Questions

Question. The co-ordinates of the point which is reflection of point \( (– 3, 5) \) in X-axis are : 
(a) \( (3, 5) \)
(b) \( (3, – 5) \)
(c) \( (– 3, – 5) \)
(d) \( (– 3, 5) \)
Answer: (c)
Sol. The co-ordinates of the point which is reflection of point \( (– 3, 5) \) in x-axis is \( (– 3, – 5) \).

Question. The distance of the point \( (– 3, 4) \) from the X-axis is :
(a) 3
(b) \( – 3 \)
(c) 4
(d) 5
Answer: (d)
Sol. The given point is \( (– 3, 4) \).
Thus, the distance = \( \sqrt{(-3 - 0)^2 + (4 - 0)^2} = 5 \) units

Question. The point on the X-axis which is equidistant from the points \( (– 1, 0) \) and \( (5, 0) \) is :
(a) \( (2, 0) \)
(b) \( (0, 2) \)
(c) \( (3, 0) \)
(d) \( (2, 2) \)
Answer: (a)
Sol. Given, \( A = (– 1, 0) \) and \( B = (5, 0) \).
Let the point on the X-axis be \( P(x, 0) \).
Thus, \( AP = PB \)
\( \Rightarrow \sqrt{(x+1)^2 + (0-0)^2} = \sqrt{(5-x)^2 + (0-0)^2} \)
\( \Rightarrow (x + 1)^2 = (5 – x)^2 \)
\( \Rightarrow x + 1 = 5 – x \)
\( \Rightarrow 2x = 4 \)
\( \Rightarrow x = 2 \)
Thus, the point is \( (2, 0) \).

Question. 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 :
(a) I
(b) II
(c) III
(d) IV
Answer: (d)
Sol. Let the coordinates of the point \( P \) be \( (x, y) \).
Given, \( A = (2, – 5), B = (5, 2) \) and \( m : n = 2 : 3 \)
Now, \( P_x = \frac{2(5) + 3(2)}{2 + 3} \) and \( P_y = \frac{2(2) + 3(-5)}{2 + 3} \)
\( \Rightarrow P_x = \frac{10 + 6}{5} \) and \( P_y = \frac{4 - 15}{5} \)
\( \Rightarrow x = \frac{16}{5} \) and \( y = -\frac{11}{5} \)
\( \Rightarrow P = \left( \frac{16}{5}, -\frac{11}{5} \right) \)
Thus, \( P \) lies in quadrant IV.

Question. If the point \( P(6, 2) \) divides the line segment joining \( A(6, 5) \) and \( B(4, y) \) in the ratio \( 3 : 1 \), then the value of \( y \) is : 
(a) 4
(b) 3
(c) 2
(d) 1
Answer: (d)
Sol. Using section formula,
\( y \)-coordinate of \( P = \frac{3 \times y + 1 \times 5}{3 + 1} \)
\( \Rightarrow 2 = \frac{3y + 5}{4} \)
\( \Rightarrow 8 = 3y + 5 \)
\( \Rightarrow 3y = 3 \)
\( \Rightarrow y = 1 \)

Fill in the Blanks

Question. _________ is the point of intersection of the coordinate axes.
Answer: Origin

Question. In the second quadrant, for a point, the abscissa is _______ and the ordinate is ________ .
Answer: Negative, positive

Question. The distance between the points \( (x, y) \) and \( (0, 0) \) is ________ units.
Answer: \( \sqrt{x^2 + y^2} \)

Question. If three points are collinear, then the area of triangle formed by them is _______
Answer: Zero

True/False

Question. The coordinates of a point on X-axis is of the form \( (x, 0) \).
Answer: True

Question. A quadrant is one fourth part of a plane divided by the coordinate axes.
Answer: True

Question. The coordinates of the mid-point of the line segment joining \( (1, 2) \) and \( (3, – 4) \) is \( (– 1, 1) \).
Answer: False
By mid-point formula, \( \left( \frac{1+3}{2}, \frac{2+(-4)}{2} \right) = \left( \frac{4}{2}, \frac{-2}{2} \right) = (2, -1) \)

Very Short Answer Type Questions

Question. \( ABCD \) is a rectangle whose vertices are \( B(4, 0), C(4, 3) \) and \( D(0, 3) \). Find the length of one of its diagonals.
Answer: In rectangle \( ABCD \), \( B(4, 0), C(4, 3) \) and \( D(0, 3) \) are the coordinates. Thus, \( BD \) is the diagonal. Hence, Length of \( BD = \sqrt{(4 - 0)^2 + (0 - 3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \) units.

Question. Find the distance of a point \( P(x, y) \) from the origin. [2018]
Answer: Distance between \( (x, y) \) and \( (0, 0) \) \( \Rightarrow \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2} \).

Question. Find the distance of the point \( (-8, 6) \) from the origin.
Answer: The given point is \( (-8, 6) \). We know that, Distance \( = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \). Thus, the distance \( = \sqrt{(-8 - 0)^2 + (6 - 0)^2} = 10 \) units.

Question. If \( A \) and \( B \) are the points \( (-6, 7) \) and \( (-1, -5) \) respectively, then find the distance \( 2AB \).
Answer: Given, \( A = (-6, 7) \) and \( B = (-1, -5) \). By distance formula, \( AB = \sqrt{(-6 + 1)^2 + (7 + 5)^2} = \sqrt{(-5)^2 + (12)^2} = \sqrt{25 + 144} = 13 \). Thus, \( 2AB = 2 \times 13 = 26 \) units.

Question. In the given figure, \( P(5, -3) \) and \( Q(3, y) \) are the points of trisection of the line segment joining \( A(4, 7) \) and \( B(1, -5) \). Then find \( y \).
Answer: Since \( P \) and \( Q \) are the points of trisection, thus, \( PQ = QB \)
\( \Rightarrow \sqrt{(5 - 3)^2 + (-3 - y)^2} = \sqrt{(3 - 1)^2 + (y + 5)^2} \)
\( \Rightarrow (2)^2 + (3 + y)^2 = (2)^2 + (y + 5)^2 \)
\( \Rightarrow 4 + 9 + 6y + y^2 = 4 + y^2 + 10y + 25 \)
\( \Rightarrow 6y + 13 = 29 + 10y \)
\( \Rightarrow 4y = -16 \)
\( \Rightarrow y = -4 \).

Question. Write the coordinates of a point on the X-axis which is equidistant from the points \( A(-2, 0) \) and \( B(6, 0) \). 
Answer: Given, \( A = (-2, 0) \) and \( B = (6, 0) \). Let the point on the X-axis be \( P(x, 0) \). It is given that \( AP = PB \)
\( \Rightarrow \sqrt{(x + 2)^2 + (0 - 0)^2} = \sqrt{(6 - x)^2 + (0 - 0)^2} \)
\( \Rightarrow (x + 2)^2 = (6 - x)^2 \)
\( \Rightarrow x + 2 = 6 - x \)
\( \Rightarrow 2x = 4 \)
\( \Rightarrow x = 2 \)
Thus, the point is \( (2, 0) \).

Question. Find the value of \( y \) for which the distance between the points \( A(3, -1) \) and \( B(11, y) \) is 10 units.
Answer: Given : \( A = (3, -1), B = (11, y) \) and \( AB = 10 \) units. Now, \( 10 = \sqrt{(11 - 3)^2 + (y + 1)^2} \)
\( \Rightarrow 100 = 64 + 1 + 2y + y^2 \)
\( \Rightarrow y^2 + 2y - 35 = 0 \)
\( \Rightarrow y^2 + 7y - 5y - 35 = 0 \)
\( \Rightarrow y(y + 7) - 5(y + 7) = 0 \)
\( \Rightarrow (y + 7) (y - 5) = 0 \)
\( \Rightarrow y = 5, -7 \)
Thus the value of \( y \) is 5 or \( -7 \).

Question. Find a relation between \( x \) and \( y \) such that the point \( P(x, y) \) is equidistant from the points \( A(1, 4) \) and \( B(-1, 2) \). 
Answer: Given : \( A = (1, 4), B = (-1, 2) \) and \( P = (x, y) \). Also, \( AP = PB \)
\( \Rightarrow \sqrt{(x - 1)^2 + (y - 4)^2} = \sqrt{(x + 1)^2 + (y - 2)^2} \)
\( \Rightarrow (x - 1)^2 + (y - 4)^2 = (x + 1)^2 + (y - 2)^2 \)
\( \Rightarrow (x - 1)^2 - (x + 1)^2 = (y - 2)^2 - (y - 4)^2 \)
\( \Rightarrow [(x - 1 + x + 1)(x - 1 - x - 1)] = [(y - 2 + y - 4)(y - 2 - y + 4)] \)
\( \Rightarrow 2x(-2) = [(2)(2y - 6)] \)
\( \Rightarrow -4x = 4(y - 3) \)
\( \Rightarrow -x = y - 3 \)
\( \Rightarrow x + y = 3 \).

Question. Find the coordinates of any point on the Y-axis which is nearest to the point \( (-2, 5) \). 
Answer: Let the required point be \( (0, y) \). Then, by distance formula, Distance \( = \sqrt{(0 - (-2))^2 + (y - 5)^2} = \sqrt{(2)^2 + (y - 5)^2} \). Now, this distance will be minimum, if \( y - 5 = 0 \Rightarrow y = 5 \). Thus, the required point is \( (0, 5) \).

Question. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is \( (2, -3) \) and B is \( (1, 4) \). 
Answer: Let the coordinates of point A be \( (x, y) \). We know, centre of a circle is the mid-point of two ends of a diameter. \(\therefore\) By mid-point formula, \( (2, -3) = \left( \frac{x + 1}{2}, \frac{y + 4}{2} \right) \Rightarrow 2 = \frac{x+1}{2}; -3 = \frac{y+4}{2} \Rightarrow x = 4 - 1; y = -6 - 4 \Rightarrow x = 3; y = -10 \). Thus, the coordinates of point A are \( (3, -10) \).

Question. If the centroid of a triangle formed by points \( P(a, b) \), \( Q(b, c) \) and \( R(c, a) \) is at the origin, what is the value of \( a + b + c \)?
Answer: We know, Centroid = \( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \). \(\therefore (0, 0) = \left( \frac{a + b + c}{3}, \frac{b + c + a}{3} \right) \Rightarrow a + b + c = 0 \).

Short Answer Type Questions-I

Question. If the point \( (x, y) \) is equidistant from the points \( (a + b, b - a) \) and \( (a - b, a + b) \), prove that \( ax = by \).
Answer: Given : \( (x, y) \) is equidistant from \( (a + b, b - a) \) and \( (a - b, a + b) \).
\(\therefore \sqrt{[x - (a + b)]^2 + [y - (b - a)]^2} = \sqrt{[x - (a - b)]^2 + [y - (a + b)]^2}\)
Squaring both sides,
\( \{x - (a + b)\}^2 + \{y - (b - a)\}^2 = \{x - (a - b)\}^2 + \{y - (a + b)\}^2 \)
\( \Rightarrow x^2 + (a + b)^2 - 2x(a + b) + y^2 + (b - a)^2 - 2y(b - a) = x^2 + (a - b)^2 - 2x(a - b) + y^2 + (a + b)^2 - 2y(a + b) \)
\( \Rightarrow - 2xa - 2xb - 2yb + 2ya = - 2xa + 2xb - 2ya - 2yb \)
\( \Rightarrow - 4xb = - 4ya \)
\( \Rightarrow bx = ay \). Hence Proved.

Question. Find the coordinates of a point on the X-axis which is equidistant from the points \( A(2, -5) \) and \( B(-2, 9) \). 
Answer: Let the required point on X-axis be \( C(x, 0) \).
Then, according to the question, AC = BC
\( \Rightarrow \sqrt{(x - 2)^2 + (0 - (-5))^2} = \sqrt{(x - (-2))^2 + (0 - 9)^2} \)
Squaring both sides,
\( (x - 2)^2 + (5)^2 = (x + 2)^2 + (9)^2 \)
\( \Rightarrow x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81 \)
\( \Rightarrow - 8x = 56 \)
\( \Rightarrow x = - \frac{56}{8} = - 7 \)
Thus, the required point is \( (-7, 0) \).

Question. If the points \( (p, q) \), \( (m, n) \) and \( (p - m, q - n) \) are collinear, show that \( pn = qm \).
Answer: Given \( (p, q) \), \( (m, n) \) and \( (p - m, q - n) \) are collinear if the area of triangle formed by them is zero. i.e., \( \Delta = 0 \)
Thus, \( \frac{1}{2} |p[n - (q - n)] + m[(q - n) - q] + (p - m)(q - q)| = 0 \) [Correction from OCR: \( \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| = 0 \)]
\( \Rightarrow p(n - q + n) + m(q - n - q) + (p - m)(q - n) = 0 \)
\( \Rightarrow p(2n - q) + m(- n) + pq - mq - pn + mn = 0 \)
\( \Rightarrow 2pn - pq - mn + pq - mq - pn + mn = 0 \)
\( \Rightarrow pn - mq = 0 \)
\(\therefore pn = mq \). Hence Proved.

Question. Find the value of \( k \) for which the points \( (-5, 1) \), \( (1, k) \) and \( (4, -2) \) are collinear. 
Answer: Given points are collinear if the area of triangle formed by them is zero.
i.e., \( \Delta = 0 \)
\( \Rightarrow \frac{1}{2} | -5(k - (-2)) + 1(-2 - 1) + 4(1 - k) | = 0 \)
\( \Rightarrow \frac{1}{2} | -5(k + 2) + 1(-3) + 4(1 - k) | = 0 \)
\( \Rightarrow | - 5k - 10 - 3 + 4 - 4k | = 0 \)
\( \Rightarrow | - 9k - 9 | = 0 \)
\( \Rightarrow - 9k = 9 \)
\( \Rightarrow k = -1 \).

Question. If \( P(x, y) \) is any point on the line joining the points \( A(a, 0) \) and \( B(0, b) \), then show that \( \frac{x}{a} + \frac{y}{b} = 1 \).
Answer: Sol. Given, \( A = (a, 0) \) and \( B = (0, b) \). Also \( P = (x, y) \) is collinear with A and B.
Thus, area of \( \Delta PAB = 0 \)
\( \Rightarrow \frac{1}{2} |x(0 - b) + a(b - y) + 0(y - 0)| = 0 \)
\( \Rightarrow - bx + ab - ay = 0 \)
\( \Rightarrow ab = ay + bx \)
\( \Rightarrow \frac{ab}{ab} = \frac{ay}{ab} + \frac{bx}{ab} \) [Dividing both sides by ab]
\( \Rightarrow 1 = \frac{y}{b} + \frac{x}{a} \) or \( \frac{x}{a} + \frac{y}{b} = 1 \). Hence Proved.

Question. For what value of \( x \) will the points \( (x, -1) \), \( (2, 1) \) and \( (4, 5) \) lie on a line?
Answer: Given : \( A = (x, -1), B = (2, 1) \) and \( C = (4, 5) \). For A, B and C to be collinear, they must satisfy the condition of collinearity.
Thus, area of \( \Delta ABC = 0 \)
\( \Rightarrow \frac{1}{2} |x(1 - 5) + 2(5 + 1) + 4(-1 - 1)| = 0 \)
\( \Rightarrow - 4x + 12 - 8 = 0 \)
\( \Rightarrow - 4x + 4 = 0 \)
\( \Rightarrow - 4x = - 4 \)
\( \Rightarrow x = 1 \)
Thus, the points will be collinear if \( x = 1 \).

Question. Prove that the points \( (a, b + c) \), \( (b, c + a) \) and \( (c, a + b) \) are collinear.
Answer: Given : \( A = (a, b + c), B = (b, c + a) \) and \( C = (c, a + b) \).
For A, B and C to be collinear, they must satisfy the condition of collinearity.
Thus, area of \( \Delta ABC = 0 \)
\( \Rightarrow \frac{1}{2} |a(c + a - a - b) + b(a + b - b - c) + c(b + c - c - a)| = 0 \)
\( \Rightarrow a(c - b) + b(a - c) + c(b - a) = 0 \)
\( \Rightarrow ac - ab + ab - bc + bc - ac = 0 \)
\( \Rightarrow 0 = 0 \)
As both L.H.S. = R.H.S. = 0, the three points are collinear. Hence Proved.

Question. The x-coordinate of a point P is twice the y-coordinate. If P is equidistant from \( Q(2, -5) \) and \( R(-3, 6) \) then find the coordinates of P.
Answer: Let the coordinates of P be \( (x, y) \).
Now, \( x = 2y \) [Given]
Thus, \( P = (2y, y) \)
Given, \( Q = (2, -5) \) and \( R = (-3, 6) \)
Now, PQ = PR [Given]
\( \Rightarrow \sqrt{(2y - 2)^2 + (y + 5)^2} = \sqrt{(2y + 3)^2 + (y - 6)^2} \)
\( \Rightarrow 4y^2 - 8y + 4 + y^2 + 10y + 25 = 4y^2 + 12y + 9 + y^2 - 12y + 36 \)
\( \Rightarrow 29 + 2y = 45 \)
\( \Rightarrow 2y = 16 \)
\( \Rightarrow y = 8 \)
So, \( x = 2 \times 8 = 16 \)
Thus, \( P = (16, 8) \).

Question. If the vertices of a triangle are \( (1, -3) \), \( (4, p) \) and \( (-9, 7) \) and its area is 15 sq. units, find the value of \( p \).
Answer: Given, the vertices of triangle are \( (1, -3), (4, p) \) and \( (-9, 7) \) and its area is 15 sq. units.
Thus, \( 15 = \frac{1}{2} |1(p - 7) + 4(7 + 3) - 9(-3 - p)| \)
\( \Rightarrow 30 = |p - 7 + 40 + 27 + 9p| \)
\( \Rightarrow 30 = |10p + 60| \)
Either \( 30 = 10p + 60 \Rightarrow 10p = -30 \Rightarrow p = -3 \)
or \( 30 = -(10p + 60) \Rightarrow 30 = -10p - 60 \Rightarrow 10p = -90 \Rightarrow p = -9 \).
Thus, \( p = -3 \) or \( -9 \).

Question. Find the value of \( p \) for which the points \( (p + 1, 2p - 2), (p - 1, p) \) and \( (p - 3, 2p - 6) \) are collinear.
Answer: Sol. Since the points \( (p + 1, 2p - 2), (p - 1, p) \) and \( (p - 3, 2p - 6) \) are collinear, therefore we have
\( x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0 \)
\( \Rightarrow (p + 1) (p - 2p + 6) + (p - 1) (2p - 6 - 2p + 2) + (p - 3) (2p - 2 - p) = 0 \)
\( \Rightarrow (p + 1) (6 - p) + (p - 1) (-4) + (p - 3) (p - 2) = 0 \)
\( \Rightarrow -p^2 + 6p - p + 6 - 4p + 4 + p^2 - 3p - 2p + 6 = 0 \)
\( \Rightarrow -4p + 16 = 0 \)
\( \Rightarrow 4p = 16 \)
\( \Rightarrow p = 4 \)
Hence, the value of \( p \) is 4.

Question. Find a relation between \( x \) and \( y \) if the points \( A(x, y), B(-4, 6) \) and \( C(-2, 3) \) are collinear. 
Answer: Sol. Given, \( A(x, y), B(-4, 6), C(-2, 3) \)
\( x_1 = x, y_1 = y, x_2 = -4, y_2 = 6, x_3 = -2, y_3 = 3 \)
If these points are collinear, then area of triangle made by these points is 0.
\( \frac{1}{2} [x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3 (y_1 - y_2)] = 0 \)
\( \frac{1}{2} [x (6 - 3) + (-4) (3 - y) + (-2) (y - 6)] = 0 \)
\( 3x - 12 + 4y - 2y + 12 = 0 \)
\( 3x + 4y - 2y = 0 \)
\( 3x + 2y = 0 \)
\( 3x = -2y \)

Short Answer Type Questions-II

Question. If \( \left( 1, \frac{p}{3} \right) \) is the mid-point of the line segment joining the points \( (2, 0) \) and \( \left( 0, \frac{2}{9} \right) \) then show that the line \( 5x + 3y + 2 = 0 \) passes through the point \( (-1, 3p) \). [SQP 2017-18]
Answer: Sol. Since \( \left( 1, \frac{p}{3} \right) \) is the mid-point of the segment joining the points \( (2, 0) \) and \( \left( 0, \frac{2}{9} \right) \).
\( \therefore \) By mid-point formula,
\( \left( 1, \frac{p}{3} \right) = \left( \frac{2 + 0}{2}, \frac{0 + 2/9}{2} \right) \)
\( \Rightarrow \frac{p}{3} = \frac{2/9}{2} = \frac{2}{18} \)
\( \Rightarrow p = \frac{2}{18} \times 3 = \frac{1}{3} \)
So, point \( (-1, 3p) \equiv \left( -1, 3 \times \frac{1}{3} \right) \equiv (-1, 1) \)
Given line is,
\( 5x + 3y + 2 = 0 \)
\( \text{L.H.S.} = 5(-1) + 3(1) + 2 \)
\( = -5 + 3 + 2 = 0 \)
\( = \text{R.H.S.} \)
Since, \( \text{L.H.S.} = \text{R.H.S.} \)
\( \therefore \) The point \( (-1, 1) \) lies on the line \( 5x + 3y + 2 = 0 \). Hence Proved.

Question. Points \( A(-1, y) \) and \( B(5, 7) \) lie on a circle with centre \( O(2, -3y) \). Find the values of \( y \). Hence find the radius of the circle.
Answer: Sol. As \( A \) and \( B \) lie on a circle, so \( AO \) and \( BO \) form the radii.
Thus, \( AO = BO \)
\( \Rightarrow \sqrt{[2 - (-1)]^2 + (-3y - y)^2} = \sqrt{(2 - 5)^2 + (-3y - 7)^2} \)
\( \Rightarrow (2 + 1)^2 + (-4y)^2 = (-3)^2 + (-3y - 7)^2 \)
\( \Rightarrow 9 + 16y^2 = 9 + 9y^2 + 49 + 42y \)
\( \Rightarrow 7y^2 - 42y - 49 = 0 \)
\( \Rightarrow y^2 - 6y - 7 = 0 \)
\( \Rightarrow y^2 - 7y + y - 7 = 0 \)
\( \Rightarrow y(y - 7) + 1(y - 7) = 0 \)
\( \Rightarrow (y - 7) (y + 1) = 0 \)
\( \Rightarrow y = -1, 7 \)
Thus, the values of \( y \) are \( -1 \) or 7.
So, \( AO = BO = \sqrt{[2 - (-1)]^2 + (-3y - y)^2} \)
When \( y = -1 \),
\( AO = BO = \sqrt{(2 + 1)^2 + (3 + 1)^2} \)
\( \Rightarrow AO = BO = \sqrt{(3)^2 + (4)^2} \)
\( = \sqrt{9 + 16} = 5 \) units
When \( y = 7 \),
\( AO = BO = \sqrt{[2 - (-1)]^2 + (-21 - 7)^2} \)
\( = \sqrt{(3)^2 + (-28)^2} \)
\( = \sqrt{9 + 784} \)
\( = \sqrt{793} \)
\( = 28.16 \) units
Hence, the radius of the circle is either 5 units or 28.16 units.

Question. Find the ratio in which \( P(4, m) \) divides the line segment joining the points \( A(2, 3) \) and \( B(6, -3) \). Hence find \( m \). 
Answer: Sol. Points \( A(2, 3) \), \( B(6, -3) \) divided by \( P(4, m) \).
Let the ratio be \( k : 1 \).
By section formula,
\( P(4, m) = \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right) \)
\( (4, m) = \left( \frac{6k + 2}{k + 1}, \frac{-3k + 3}{k + 1} \right) \)
\( \Rightarrow \frac{6k + 2}{k + 1} = 4 \)
\( 6k + 2 = 4k + 4 \)
\( 2k = 2 \)
\( k = 1 \)
The ratio is \( 1 : 1 \).
Now, \( m = \frac{-3k + 3}{k + 1} \)
\( m = \frac{-3(1) + 3}{1 + 1} \)
\( m = 0 \)
Value of \( m \) is \( 0 \), the point is \( P(4, 0) \).

Question. 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.
Answer: Sol. Given, \( A(2, -1) \), \( B(5, -1) \), \( C(5, 6) \) and \( D(2, 6) \)
Using distance formula, \( AC = \sqrt{(5 - 2)^2 + (6 + 1)^2} \)
\( = \sqrt{(3)^2 + (7)^2} \)
\( = \sqrt{9 + 49} = \sqrt{58} \)
\( BD = \sqrt{(2 - 5)^2 + (6 + 1)^2} \)
\( = \sqrt{(-3)^2 + (7)^2} \)
\( = \sqrt{9 + 49} = \sqrt{58} \)
Thus, \( AC = BD \).
Let \( E \) be the mid-point of \( AC \) and \( F \) the mid-point of \( BD \).
Thus, \( E = \left( \frac{2 + 5}{2}, \frac{6 - 1}{2} \right) = \left( \frac{7}{2}, \frac{5}{2} \right) \)
Also, \( F = \left( \frac{5 + 2}{2}, \frac{6 - 1}{2} \right) = \left( \frac{7}{2}, \frac{5}{2} \right) \)
Hence, the diagonals \( AC \) and \( BD \) are equal and bisect each other at the point \( \left( \frac{7}{2}, \frac{5}{2} \right) \). Hence Proved.

Question. If the mid-point of the line segment joining the points \( A(3, 4) \) and \( B(k, 6) \) is \( P(x, y) \) and \( x + y - 10 = 0 \). Find the value of \( k \). 
Answer: Sol. Coordinates of the mid-point of \( A(3, 4) \) and \( B(k, 6) = \left( \frac{3 + k}{2}, \frac{4 + 6}{2} \right) = \left( \frac{3 + k}{2}, 5 \right) \)
Since \( P(x, y) \) is the mid-point of \( AB \),
\( \therefore P(x, y) = \left( \frac{3 + k}{2}, 5 \right) \)
On comparing, we get
\( x = \frac{3 + k}{2} \) and \( y = 5 \)
Now, \( x + y - 10 = 0 \)
\( \Rightarrow \frac{3 + k}{2} + 5 - 10 = 0 \)
\( \Rightarrow \frac{3 + k}{2} - 5 = 0 \)
\( \Rightarrow 3 + k - 10 = 0 \)
\( \Rightarrow k - 7 = 0 \)
\( \Rightarrow k = 7 \)

Question. Find the value of \( k \) for which the points \( (3k - 1, k - 2) \), \( (k, k - 7) \) and \( (k - 1, -k - 2) \) are collinear. 
Answer: Sol. Given points \( (3k - 1, k - 2) \), \( (k, k - 7) \) and \( (k - 1, -k - 2) \) will be collinear, if the area of triangle formed by them is zero.
i.e., \( \Delta = 0 \)
\( \Rightarrow \frac{1}{2} | (3k - 1) \{ (k - 7) - (-k - 2) \} + k \{ (-k - 2) - (k - 2) \} + (k - 1) \{ (k - 2) - (k - 7) \} | = 0 \)
\( \Rightarrow \frac{1}{2} | (3k - 1) (2k - 5) + k(-2k) + (k - 1) (5) | = 0 \)
\( \Rightarrow \frac{1}{2} | 6k^2 - 17k + 5 - 2k^2 + 5k - 5 | = 0 \)
\( \Rightarrow \frac{1}{2} | 4k^2 - 12k | = 0 \)
\( \Rightarrow 4k^2 - 12k = 0 \)
\( \Rightarrow 4k(k - 3) = 0 \)
\( \Rightarrow k = 0 \) or \( k = 3 \).

Question. Prove that the points (0, 0) (5, 5) and (– 5, 5) are the vertices of an isosceles right-angled triangle.
Answer: Let, \( A = (0, 0), B = (5, 5) \) and \( C = (– 5, 5) \).
Thus, \( AB = \sqrt{(5 - 0)^2 + (5 - 0)^2} \)
\( = \sqrt{(5)^2 + (5)^2} = \sqrt{25 + 25} \)
\( = \sqrt{50} \)
\( BC = \sqrt{(- 5 - 5)^2 + (5 - 5)^2} \)
\( = \sqrt{(- 10)^2} = 10 \)
and \( AC = \sqrt{(- 5 - 0)^2 + (5 - 0)^2} \)
\( = \sqrt{(- 5)^2 + (5)^2} = \sqrt{25 + 25} \)
\( = \sqrt{50} \)
Thus, \( AB = AC \)
Also, \( AB^2 + AC^2 = 50 + 50 = 100 = BC^2 \)
Thus, \( \Delta ABC \) is an isosceles right-angled triangle.
Hence Proved.

Question. Determine the ratio in which the line \( 2x + y – 4 = 0 \) divides the line segment joining the points A(2, – 2) and B(3, 7). 
Answer: Let the given line \( 2x + y – 4 \) intersect the segment joining the points \( A(2, – 2) \) and \( B(3, 7) \) at point \( R(x, y) \) in the ratio \( k : 1 \).
Then, by section formula
\( R(x, y) = \left( \frac{3k + 2}{k + 1}, \frac{7k + (-2)}{k + 1} \right) \)
\( \Rightarrow R(x, y) = \left( \frac{3k + 2}{k + 1}, \frac{7k - 2}{k + 1} \right) \)
Since, the point R lies on the line \( 2x + y – 4 = 0 \), therefore, it must satisfy the equation.
\( \therefore 2\left( \frac{3k + 2}{k + 1} \right) + \left( \frac{7k - 2}{k + 1} \right) - 4 = 0 \)
\( \Rightarrow 2(3k + 2) + (7k – 2) – 4(k + 1) = 0 \)
\( \Rightarrow 6k + 4 + 7k – 2 – 4k – 4 = 0 \)
\( \Rightarrow 9k – 2 = 0 \)
\( \Rightarrow k = \frac{2}{9} \)
Thus, the required ratio = \( k : 1 = \frac{2}{9} : 1 = 2 : 9 \). Ans.

Question. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).
Answer: Let \( O(x, y) \) be the centre of circle. Given points are \( A(6, – 6), B(3, –7) \) and \( C(3, 3) \).
Then, \( OA = \sqrt{(x - 6)^2 + (y + 6)^2} \);
\( OB = \sqrt{(x - 3)^2 + (y + 7)^2} \)
and \( OC = \sqrt{(x - 3)^2 + (y - 3)^2} \)
Since, each point on the circle is equidistant from centre.
\( \therefore OA = OB = OC = \text{Radius} \)
\( \Rightarrow OA = OB \)
\( \Rightarrow \sqrt{(x - 6)^2 + (y + 6)^2} = \sqrt{(x - 3)^2 + (y + 7)^2} \)
Squaring both sides, we get
\( (x – 6)^2 + (y + 6)^2 = (x – 3)^2 + (y + 7)^2 \)
\( \Rightarrow x^2 – 12x + 36 + y^2 + 12y + 36 = x^2 – 6x + 9 + y^2 + 14y + 49 \)
\( \Rightarrow – 6x – 2y = –14 \)
\( \Rightarrow 3x + y = 7 \) ...(i)
Similarly, \( OB = OC \)
\( \Rightarrow \sqrt{(x - 3)^2 + (y + 7)^2} = \sqrt{(x - 3)^2 + (y - 3)^2} \)
Squaring both sides, we get
\( (x – 3)^2 + (y + 7)^2 = (x – 3)^2 + (y – 3)^2 \)
\( \Rightarrow (y + 7)^2 = (y – 3)^2 \)
\( \Rightarrow y^2 + 14y + 49 = y^2 – 6y + 9 \)
\( \Rightarrow 20y = – 40 \)
\( \Rightarrow y = – 2 \)
Putting \( y = –2 \) in equation (i), we get
\( 3x – 2 = 7 \Rightarrow x = 3 \)
Hence, the coordinates of the centre of circle are (3, –2). Ans.

Question. If the coordinates of two points are A(3, 4) and B(5, – 2) and a point P(x, 5) is such that PA = PB, then find the area of triangle PAB. 
Answer: Given : \( PA = PB \)
\( \therefore \) By distance formula,
\( \sqrt{(x - 3)^2 + (5 - 4)^2} = \sqrt{(x - 5)^2 + (5 - (-2))^2} \)
Squaring both sides,
\( (x – 3)^2 + (1)^2 = (x – 5)^2 + (7)^2 \)
\( \Rightarrow x^2 – 6x + 9 + 1 = x^2 – 10x + 25 + 49 \)
\( \Rightarrow 4x = 64 \)
\( \Rightarrow x = 16 \)
\( \Rightarrow \) Coordinates of \( P = (16, 5) \)
Now, Area of \( \Delta PAB \)
\( = \frac{1}{2} | 16(4 - (- 2)) + 3(- 2 - 5) + 5(5 - 4) | \)
\( = \frac{1}{2} | 16(6) + 3(- 7) + 5(1) | \)
\( = \frac{1}{2} | 96 - 21 + 5 | = \frac{1}{2} | 80 | = 40 \) sq. units. Ans.