CBSE Class 10 Mathematics Coordinate Geometry VBQs Set C

Multiple Choice Questions

Question. The point on the x-axis which is equidistant from (– 4, 0) and (10, 0) is:
(a) (7, 0)
(b) (5, 0)
(c) (0, 0)
(d) (3, 0) 
Answer: (d)
Explanation: Since, both the given points are on the x-axis, the mid-point \( \left( \frac{-4+10}{2}, \frac{0+0}{2} \right) \), i.e., (3, 0) lies on x-axis and is equidistant from (–4, 0) and (10, 0)

Question. The centre of a circle whose end points of a diameter are (– 6, 3) and (6, 4) are:
(a) (8, – 1)
(b) (4, 7)
(c) \( \left( 0, \frac{7}{2} \right) \)
(d) \( \left( 4, \frac{7}{2} \right) \) 
Answer: (c)
Explanation: Since, the centre of a circle is the mid-point of its diameter, the centre is \( \left( \frac{-6+6}{2}, \frac{3+4}{2} \right) \), i.e., \( \left( 0, \frac{7}{2} \right) \)

Question. The distance between the points (m, – n) and (– m, n) is:
(a) \( \sqrt{m^2 + n^2} \)
(b) \( m + n \)
(c) \( 2\sqrt{m^2 + n^2} \)
(d) \( \sqrt{2m^2 + 2n^2} \) 
Answer: (c)
Explanation: Distance between (m, –n) and (–m, n) is
\( = \sqrt{(-m - m)^2 + (n - (-n))^2} \) (By distance formula)
\( = \sqrt{4m^2 + 4n^2} \)
\( = 2\sqrt{m^2 + n^2} \)

Question. The point which divides the line segment joining the points (7, –6) and (3, 4) in the ratio 1:2 internally, lies in the:
(a) I quadrant
(b) II quadrant
(c) III quadrant
(d) IV quadrant 
Answer: (d)
Explanation: We know that if \( P(x, y) \) divides the line segment joining \( A(x_1, y_1) \) and \( B(x_2, y_2) \) internally in the ratio \( m:n \), then
\( x = \frac{mx_2 + nx_1}{m+n} \) and \( y = \frac{my_2 + ny_1}{m+n} \)
Given that \( x_1 = 7, y_1 = -6, x_2 = 3, y_2 = 4, m = 1, n = 2 \)
\( \therefore x = \frac{1(3) + 2(7)}{1+2} = \frac{3 + 14}{3} = \frac{17}{3} \)
\( y = \frac{1(4) + 2(-6)}{1+2} = \frac{4 - 12}{3} = -\frac{8}{3} \)
As x-coordinate is positive and y-coordinate is negative:
\( \therefore (x, y) = \left( \frac{17}{3}, -\frac{8}{3} \right) \) lies is the IV quadrant

Question. The distance between the points \( (a \cos \theta + b \sin \theta, 0) \) and \( (0, a \sin \theta - b \cos \theta) \), is
(a) \( a^2 + b^2 \)
(b) \( a^2 - b^2 \)
(c) \( \sqrt{a^2 + b^2} \)
(d) \( \sqrt{a^2 - b^2} \)
Answer: (c)
Explanation: Distance between the given points
\( = \sqrt{(0 - (a \cos \theta + b \sin \theta))^2 + (a \sin \theta - b \cos \theta - 0)^2} \)
\( = \sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \cos \theta \sin \theta + a^2 \sin^2 \theta + b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta} \)
\( = \sqrt{a^2 + b^2} \)

Question. The point which lies on the perpendicular bisector of the line segment joining point A (–2, –5) and B (2, 5) is:
(a) (0, 0)
(b) (0, –1)
(c) (–1, 0)
(d) (1, 0) 
Answer: (a)
Explanation: Let \( P(\alpha, \beta) \) be a point which lies on the perpendicular bisector of the line segment AB.
Then, \( PA = PB \) or \( PA^2 = PB^2 \)
\( \Rightarrow (\alpha + 2)^2 + (\beta + 5)^2 = (\alpha - 2)^2 + (\beta - 5)^2 \)
\( \Rightarrow \alpha^2 + 4\alpha + 4 + \beta^2 + 10\beta + 25 = \alpha^2 - 4\alpha + 4 + \beta^2 - 10\beta + 25 \)
\( \Rightarrow 8\alpha + 20\beta = 0 \)
This is true only when \( \alpha = 0 \) and \( \beta = 0 \)
[From the given four options]
So, (0, 0) lies on the perpendicular bisector.

Question. The fourth vertex D of a parallelogram ABCD whose three vertices are A (–2, 3), B (6, 7) and C (8, 3) is:
(a) (0, 1)
(b) (0, –1)
(c) (–1, 0)
(d) (1, 0) 
Answer: (b)
Explanation: It is given that ABCD is a parallelogram with vertices A (–2, 3), B (6, 7) and C (8, 3). Let fourth vertex be D (x, y). We know that diagonals AC and BD will bisect each other.
Midpoint of diagonal AC \( (x_1, y_1) \)
\( = \left( \frac{-2+8}{2}, \frac{3+3}{2} \right) = \left( \frac{6}{2}, \frac{6}{2} \right) = (3, 3) \)
Midpoint of diagonal BD \( (x_2, y_2) = \left( \frac{x+6}{2}, \frac{y+7}{2} \right) \)
But the two midpoints are the same. So,
\( \frac{x+6}{2} = 3 \) and \( \frac{y+7}{2} = 3 \)
\( \Rightarrow x + 6 = 6 \) and \( y + 7 = 6 \)
\( \Rightarrow x = 0 \) and \( y = –1 \)
Hence, the fourth vertex D (x, y) = (0, –1).

Question. If the point P(k, 0) divides the line segment joining the points A(2, –2) and B(–7, 4) in the ratio 1 : 2, then the value of k is
(a) 1
(b) 2
(c) –2
(d) –1 
Answer: (d)
Explanation:
Here, \( P(k, 0) = \left( \frac{1(-7) + 2(2)}{3}, \frac{1(4) + 2(-2)}{3} \right) \), i.e., \( \left( \frac{-7+4}{3}, \frac{4-4}{3} \right) = (-1, 0) \) [By Section Formula]
\( \Rightarrow k = –1 \)

Question. The distance of the point P(-3, -4) from the x-axis (in units) is:
(a) 3
(b) –3
(c) 4
(d) 5
Answer: (c) 

Question. If the point P(2, 1) lies on the line segment joining points A(4, 2) and B(8, 4), then:
(a) \( AP = \frac{1}{3}AB \)
(b) \( AP = PB \)
(c) \( PB = \frac{1}{3}AB \)
(d) \( AP = \frac{1}{2}AB \) 
Answer: (d)
Explanation: It is given that P(2, 1) lies on the line segment joining the points A(4, 2) and B(8, 4).
We know that distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \),
\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Distance between A(4, 2) and P(2, 1)
\( AP = \sqrt{(2 - 4)^2 + (1 - 2)^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \)
Distance between P(2, 1) and B(8, 4)
\( PB = \sqrt{(8 - 2)^2 + (4 - 1)^2} = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \)
Distance between A(4, 2) and B(8, 4)
\( AB = \sqrt{(8 - 4)^2 + (4 - 2)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \)
\( \therefore AB = 2\sqrt{5} = 2(AP) \Rightarrow AP = \frac{AB}{2} \)

Question. If \( A \left( \frac{m}{3}, 5 \right) \) is the mid-point of the line segment joining the points Q(–6, 7) and R(–2, 3), then the value of m is:
(a) –12
(b) –4
(c) 12
(d) –6 
Answer: (a)
Explanation: Mid point is \( \frac{m}{3} = \frac{-6 - 2}{2} \)
\( \Rightarrow \frac{m}{3} = -4 \)
\( \Rightarrow m = -12 \)

Question. The perimeter of a triangle ABC with vertices A (0, 4), B (0, 0) and C (3, 0) is:
(a) 5 units
(b) 11 units
(c) 12 units
(d) \( (7 + \sqrt{5}) \) units
Answer: (c)
Explanation: Perimeter of triangle \( = AB + BC + CA \)
\( = \sqrt{(0-0)^2 + (4-0)^2} + \sqrt{(3-0)^2 + (0-0)^2} + \sqrt{(3-0)^2 + (0-4)^2} \)
\( = \sqrt{16} + \sqrt{9} + \sqrt{25} \)
\( = 4 + 3 + 5 \)
\( = 12 \text{ units} \)

Question. If \( P \left( \frac{a}{3}, 4 \right) \) is the midpoint of the line segment joining the points Q(–6, 5) and R(–2, 3), then the value of a is:
(a) –4
(b) –12
(c) 12
(d) –6
Answer: (b)
Explanation: It is given that \( P \left( \frac{a}{3}, 4 \right) \) is the midpoint of the line segment joining the points Q (–6, 5) and R (–2, 3).
We know that midpoint \( (x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
Midpoint of QR \( = P \left( \frac{-6 - 2}{2}, \frac{5 + 3}{2} \right) = \left( \frac{-8}{2}, \frac{8}{2} \right) = P(-4, 4) \)
But mid-point is given as \( P \left( \frac{a}{3}, 4 \right) \) comparing the two, we get
\( P \left( \frac{a}{3}, 4 \right) = (-4, 4) \)
\( \Rightarrow \frac{a}{3} = -4 \Rightarrow a = -12 \)
Hence, the required value of a is –12.

Question. The perpendicular bisector of the line segment joining the points A(1, 5) and B(4, 6) cuts the y-axis at:
(a) (0, 13)
(b) (0, –13)
(c) (0, 12)
(d) (13, 0)
Answer: (a)
Explanation: Let the perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cut the y-axis at P. Then point P will be of the form (0, b) as any point on the y-axis will have x-coordinates 0.
We know that AP = PB
\( \sqrt{(0-1)^2 + (b-5)^2} = \sqrt{(4-0)^2 + (6-b)^2} \)
\( \Rightarrow 1 + (b-5)^2 = 16 + (6-b)^2 \)
\( \Rightarrow 1 + (b-5)^2 = 16 + (6-b)^2 \)
\( \Rightarrow 1 + b^2 + 25 - 10b = 16 + 36 + b^2 - 12b \)
\( \Rightarrow 12b - 10b = 52 - 26 \)
\( \Rightarrow 2b = 26 \Rightarrow b = 13 \)
Hence, point P is (0, 13).

Question. A circle drawn with origin as the centre passes through \( \left( \frac{13}{2}, 0 \right) \). The point which does not lie in the interior of the circle is:
(a) \( \left( -\frac{3}{4}, 1 \right) \)
(b) \( \left( 2, \frac{7}{3} \right) \)
(c) \( \left( 5, -\frac{1}{2} \right) \)
(d) \( \left( -6, \frac{5}{2} \right) \)
Answer: (d)
Explanation: It is given that centre of the circle is origin O (0, 0) and it passes through \( \left( \frac{13}{2}, 0 \right) \).
\( \Rightarrow \) Radius of circle \( = \) Distance between (0, 0) and \( \left( \frac{13}{2}, 0 \right) \).
\( = \sqrt{\left( \frac{13}{2} - 0 \right)^2 + (0 - 0)^2} = \sqrt{\left( \frac{13}{2} \right)^2} = \frac{13}{2} = 6.5 \)
We know that the point which does not lie in the interior of circle will be at a distance greater than the radius from the centre.
(A) Distance between (0, 0) and \( \left( -\frac{3}{4}, 1 \right) \)
\( = \sqrt{\left( -\frac{3}{4} - 0 \right)^2 + (1 - 0)^2} = \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{25}{16}} = \frac{5}{4} = 1.25 \)
Clearly, \( 1.25 < 6.5 \)
\( \Rightarrow \) Point \( \left( -\frac{3}{4}, 1 \right) \) lies in the interior to the circle.
(B) Distance between (0, 0) and \( \left( 2, \frac{7}{3} \right) \)
\( = \sqrt{(2 - 0)^2 + \left( \frac{7}{3} - 0 \right)^2} = \sqrt{4 + \frac{49}{9}} = \sqrt{\frac{36+49}{9}} = \frac{\sqrt{85}}{3} = \frac{9.22}{3} = 3.1 \)
clearly \( 3.1 < 6.5 \)
\( \Rightarrow \) Point \( \left( 2, \frac{7}{3} \right) \) lies in the interior of the circle.
(C) Distance between (0, 0) and \( \left( 5, -\frac{1}{2} \right) \)
\( = \sqrt{(5 - 0)^2 + \left( -\frac{1}{2} - 0 \right)^2} = \sqrt{25 + \frac{1}{4}} = \sqrt{\frac{101}{4}} = \frac{10.04}{2} = 5.02 \)
Clearly, \( 5.02 < 6.5 \).
\( \Rightarrow \) Point \( \left( 5, -\frac{1}{2} \right) \) lies in the interior of the circle.
(D) Distance between (0, 0) and \( \left( -6, \frac{5}{2} \right) \)
\( = \sqrt{(-6 - 0)^2 + \left( \frac{5}{2} - 0 \right)^2} = \sqrt{36 + \frac{25}{4}} = \sqrt{\frac{144+25}{4}} = \frac{\sqrt{169}}{2} = \frac{13}{2} = 6.5 \)
\( = 6.5 = \text{radius} \)
So the point \( \left( -6, \frac{5}{2} \right) \) lies on the circle and not in the interior.

Fill in the Blanks

Fill in the blanks/tables with suitable information:

Question. AOBC is a rectangle whose three vertices are A(0, –3), O(0, 0) and B(4, 0). The length of its diagonal is .............................. . 
Answer: 5
Explanation: In a rectangle, both the diagonals are of equal lengths. So, by the distance formula,
\( AB = \sqrt{(4-0)^2 + (0+3)^2} \)
\( = \sqrt{16+9} \)
\( = \sqrt{25} \)
\( = 5 \)
\( \therefore AB = OC = 5 \) units

Question. The centroid of the triangle whose vertices are (4, – 8), (– 9, 7) and (8, 13) is .............................. .
Answer: (1, 4)
Explanation: Centroid of triangle having vertices \( (x_1, y_1), (x_2, y_2) \) and \( (x_3, y_3) \) is given by
\( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)
Required centroid \( = \left( \frac{4 - 9 + 8}{3}, \frac{-8 + 7 + 13}{3} \right) \)
\( = \left( \frac{3}{3}, \frac{12}{3} \right) = (1, 4) \)

Question. The ratio in which x-axis divides the line segment joining the point (2, 3) and (4, – 8) is .............................. .
Answer: \( \frac{3}{8} \)
Explanation: Let (x, 0) be the point on x-axis divides the segment joining (2, 3) and (4, – 8)
Let ratio be k : 1
\( x = \frac{4k + 2}{k + 1}, 0 = \frac{-8k + 3}{k + 1} \)
\( 0 = \frac{-8k + 3}{k + 1} \Rightarrow 8k = 3 \Rightarrow k = \frac{3}{8} \)

Question. The mid-point of the line segment AB is (4, 0). If the co-ordinates of point A is (3,–2), then co-ordinates of point B is .............................. .
Answer: B (3, 2)
Explanation: Let coordinates of B be (x, y)
Here A (3, – 2), B (x, y) and mid point (4, 0)
\( (4, 0) = \left( \frac{3 + x}{2}, \frac{-2 + y}{2} \right) \)
On comparing, we get
\( 4 = \frac{3 + x}{2} \) and \( 0 = \frac{-2 + y}{2} \)
\( \Rightarrow 3 + x = 6 \) and \( -2 + y = 0 \)
\( \Rightarrow x = 3 \) and \( y = 2 \)
\( \therefore \) Coordinates of B is (3, 2)

Question. Distance of a point (–24, 7) from the origin (in units) is .............................. .
Answer: 25 units
Explanation: Distance between (– 24, 7) and (0, 0)
\( d = \sqrt{(-24 - 0)^2 + (7 - 0)^2} \)
\( = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} \)
\( = \sqrt{625} = 25 \) units

Question. If P(–1, 1) is the mid-point of the line segment joining the points A(–3, b) and B(1, b + 4) then b = ...............................
Answer: –1
Explanation: A (– 3, b) and B (1, b + 4)
Mid-point of AB = (– 1, 1)
\( \therefore (-1, 1) = \left( \frac{-3 + 1}{2}, \frac{b + (b + 4)}{2} \right) \)
\( \Rightarrow 1 = \frac{2b + 4}{2} \)
\( \Rightarrow 2b + 4 = 2 \)
\( \Rightarrow 2b = -2 \Rightarrow b = -1 \)

Write True or False

Question. \(\Delta ABC\) with vertices A(–2, 0), B(2, 0) and C(0, 2) is similar to \(\Delta DEF\) with vertices D(–4, 0), E(4, 0) and F(0, 4). 
Answer: True.
We know that distance between the points \( (x_1, y_1) \) and \( (x_2, y_2), d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Distance between A (–2, 0) and B (2, 0),
\( AB = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4 \)
Distance between B (2, 0) and C (0, 2),
\( BC = \sqrt{(0 - 2)^2 + (2 - 0)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \)
Distance between C (0, 2) and A (–2, 0),
\( CA = \sqrt{(-2 - 0)^2 + (0 - 2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \)
Distance between D (–4, 0) and E (4, 0),
\( DE = \sqrt{(4 - (-4))^2 + (0 - 0)^2} = \sqrt{8^2} = 8 \)
Distance between E(4, 0) and F (0, 4),
\( EF = \sqrt{(0 - 4)^2 + (4 - 0)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \)
Distance between F(0, 4) and D (–4, 0),
\( FD = \sqrt{(-4 - 0)^2 + (0 - 4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \)
Now, \( \frac{AB}{DE} = \frac{4}{8} = \frac{1}{2}, \frac{BC}{EF} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2}, \frac{CA}{FD} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2} \)
\( \Rightarrow \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} \)
We see that sides of \( \Delta ABC \) and \( \Delta FDE \) are proportional. Hence \( \Delta ABC \) is similar to \( \Delta DEF \).

Question. Points A(4, 3), B(6, 4), C(5, –6) and D(–3, 5) are the vertices of a parallelogram. 
Answer: False.
We know that opposite sides of parallelogram are equal in length.
\( AB = \sqrt{(6-4)^2 + (4-3)^2} = \sqrt{2^2 + 1^2} = \sqrt{5} \)
\( BC = \sqrt{(5-6)^2 + (-6-4)^2} = \sqrt{(-1)^2 + (-10)^2} = \sqrt{1 + 100} = \sqrt{101} \)
\( CD = \sqrt{(-3-5)^2 + (5-(-6))^2} = \sqrt{(-8)^2 + (11)^2} = \sqrt{64 + 121} = \sqrt{185} \)
\( DA = \sqrt{(4-(-3))^2 + (3-5)^2} = \sqrt{7^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53} \)
As \( AB \neq CD \) and \( BC \neq DA \).
Hence, the given vertices are not vertices of a parallelogram.

Question. Point P(5, –3) is one of the two points of trisection of the line segment joining points A(7, –2) and B(1, –5). 
Answer: True.
Let point P(5, –3) divide the line segment joining the points A(7, –2) and B(1, –5) in the ratio m:1 internally.
Using section formula, Coordinate of point P will be
\( \left[ \frac{m(1)+1(7)}{m+1}, \frac{m(-5)+1(-2)}{m+1} \right] = \left[ \frac{m+7}{m+1}, \frac{-5m-2}{m+1} \right] \)
According to the question
\( (5, -3) = \left( \frac{m+7}{m+1}, \frac{-5m-2}{m+1} \right) \)
\( \Rightarrow 5 = \frac{m+7}{m+1} \) and \( -3 = \frac{-5m-2}{m+1} \)
\( \Rightarrow 5(m + 1) = (m + 7) \) and \( -3(m + 1) = -5m - 2 \)
\( \Rightarrow 5m + 5 = m + 7 \) and \( -3m - 3 = -5m - 2 \)
\( \Rightarrow 4m = 2 \) and \( 2m = 1 \)
\( \Rightarrow m = \frac{1}{2} \)
Hence, point P divides the line segment AB in the ratio 1:2. Thus point P is the point of trisection of AB.

Question. Point P(–2, 4) lies on a circle of radius 6 and centre C(3, 5). 
Answer: False.
We know that if the distance between the centre and point P is equal to the radius, then the point lies on the circle. Distance between centre C(3, 5) and point P(–2, 4):
\( PC = \sqrt{(-2 - 3)^2 + (4 - 5)^2} = \sqrt{(-5)^2 + (-1)^2} \)
\( = \sqrt{25 + 1} = \sqrt{26} \)
Clearly, \( PC \neq \text{radius (6)} \)
Hence, point P (–2, 4) does not lie on the circle with centre C (3, 5) and radius 6.

Question. The points A (–1, –2), B (4, 3), C (2, 5) and D (–3, 0) in that order from a rectangle. 
Answer: True.
We know that opposite side of a rectangle are equal and also its diagonals are equal and bisect each other.
A (–1, –2), B (4, 3), C (2, 5) and D (–3, 0)
Distance between A (–1, –2) and B (4, 3)
\( AB = \sqrt{(4 - (-1))^2 + (3 - (-2))^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2} \)
Distance between B (4, 3) and C (2, 5)
\( BC = \sqrt{(2 - 4)^2 + (5 - 3)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = 2\sqrt{2} \)
Distance between C (2, 5) and D (–3, 0)
\( CD = \sqrt{(-3 - 2)^2 + (0 - 5)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{50} = 5\sqrt{2} \)
Distance between A (–1, –2) and D (–3, 0)
\( AD = \sqrt{(-3 - (-1))^2 + (0 - (-2))^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = 2\sqrt{2} \)
Distance between A (–1, –2) and C (2, 5)
\( AC = \sqrt{(2 - (-1))^2 + (5 - (-2))^2} = \sqrt{3^2 + 7^2} = \sqrt{9 + 49} = \sqrt{58} \)
Distance between B (4, 3) and D (–3, 0)
\( BD = \sqrt{(-3 - 4)^2 + (0 - 3)^2} = \sqrt{(-7)^2 + (-3)^2} = \sqrt{49 + 9} = \sqrt{58} \)
Clearly, AB = CD, AD = BC and AC = BD
i.e., opposite sides are equal and diagonals are also equal.
Hence, points A (–1, –2), B (4, 3), C (2, 5) and D (–3, 0) form a rectangle.

Very Short Answer Type Questions

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

Question. Find the distance between the points (a, b) and (–a, –b). 
Answer: Given: points are A (a, b) and B (– a, – b). By the distance formula:
Required distance \( = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Here, \( x_1 = a, y_1 = b, x_2 = -a, y_2 = -b \)
\( AB = \sqrt{(-a - a)^2 + (-b - b)^2} \)
\( = \sqrt{(-2a)^2 + (-2b)^2} \)
\( = \sqrt{4a^2 + 4b^2} \)
\( = 2\sqrt{a^2 + b^2} \) units
Hence, the distance between the given points is \( 2\sqrt{a^2 + b^2} \) units.

SHORT ANSWER (SA-I) Type Questions

Question. Find the value of ‘a’ so that the point (3, a) lies on the line represented by 2x – 3y = 5. 
Answer: Given: line is 2x – 3y = 5. If point (3, a) lies on the given line, then this point will satisfy the equation.
\( 2(3) – 3(a) = 5 \)
\( \Rightarrow 6 - 3a = 5 \)
\( \Rightarrow 3a = 6 - 5 = 1 \)
\( \Rightarrow a = \frac{1}{3} \)
Hence, the value of a is \( \frac{1}{3} \).

Question. The mid-point of the line segment joining A(2a, 4) and B(–2, 3b) is (1, 2a +1). Find the value of a and b. 
Answer: Let, P be the mid point of the line AB. Coordinates of A are (2a, 4); B are (–2, 3b) and P are (1, 2a +1).
By the mid-point formula: Coordinates of the mid point \( (x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
Here \( x_1 = 2a, y_1 = 4, x_2 = -2, y_2 = 3b, x = 1, y = 2a + 1 \)
\( 1 = \frac{2a + (-2)}{2} \Rightarrow 2 = 2a - 2 \Rightarrow 2a = 4 \Rightarrow a = 2 \)
And \( 2a + 1 = \frac{4 + 3b}{2} \)
Put \( a = 2 \):
\( 2(2) + 1 = \frac{4 + 3b}{2} \Rightarrow 5 = \frac{4 + 3b}{2} \Rightarrow 10 = 4 + 3b \Rightarrow 3b = 6 \Rightarrow b = 2 \)
Hence, the value of a = 2 and b = 2.

Question. Determine the ratio in which the line y – x + 2 = 0 divides the line segment joining the points (3, –1) and (8, 9). 
Answer: Let, line y – x + 2 = 0 divides the points (3, –1) and (8, 9) in ratio k : 1 at point P.
x coordinate of the point \( = \frac{8k+3}{k+1} \) [\(\because \frac{mx_2+nx_1}{m+n}\)]
y coordinate of the point \( = \frac{9k-1}{k+1} \) [\(\because \frac{my_2+ny_1}{m+n}\)]
coordinates of the point P are \( \left( \frac{8k+3}{k+1}, \frac{9k-1}{k+1} \right) \)
Also, this point lies on line y – x + 2 = 0
\( \left( \frac{9k-1}{k+1} \right) - \left( \frac{8k+3}{k+1} \right) + 2 = 0 \)
\( \Rightarrow 9k – 1 – 8k – 3 + 2(k + 1) = 0 \)
\( \Rightarrow 9k – 1 – 8k – 3 + 2k + 2 = 0 \)
\( \Rightarrow 3k – 2 = 0 \)
\( \Rightarrow k = \frac{2}{3} \)
Hence, line divides in ratio 2 : 3 internally.

Question. In what ratio does the point \( P(– 4, 6) \) divide the line segment joining the points \( A(– 6, 10) \) and \( B(3, – 8) \)? 
Answer: Let, the ratio in which \( P \) divides line \( AB \) be \( k : 1 \).
By section formula, \( P(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \)
Here, \( x = – 4, y = 6, x_1 = – 6, y_1 = 10, x_2 = 3, y_2 = – 8, m = k, n = 1 \)
Now \( P(– 4, 6) = \left( \frac{k \times 3 + 1 \times (-6)}{k + 1}, \frac{k \times (-8) + 1 \times 10}{k + 1} \right) \)
\( \therefore -4 = \frac{3k - 6}{k + 1} \) or \( 6 = \frac{-8k + 10}{k + 1} \)
\( \Rightarrow -4k - 4 = 3k - 6, 6k + 6 = -8k + 10 \)
\( \Rightarrow -7k = -2, 14k = 4 \)
\( \Rightarrow k = \frac{2}{7} \)
Hence, the required ratio is \( 2 : 7 \).

Question. Find the ratio in which the point \( (– 3, k) \) divides the line-segment joining the points \( (–5, –4) \) and \( (–2, 3) \). Also, find the value of \( k \). 
Answer: Let, the ratio in which point \( P \) divides line \( AB \) be \( m : 1 \).
Here, the coordinates are: \( A(– 5, – 4), B(– 2, 3) \) and \( P(– 3, k) \)
By the section formula, \( P(x, y) = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right) \)
\( (– 3, k) = \left( \frac{m(-2) + 1(-5)}{m + 1}, \frac{m(3) + 1(-4)}{m + 1} \right) \)
\( \therefore -3 = \frac{-2m - 5}{m + 1} \)
\( \Rightarrow -3(m + 1) = -2m - 5 \)
\( \Rightarrow -3m - 3 = -2m - 5 \)
\( \Rightarrow -m = -2 \Rightarrow m = 2 \)
The required ratio is \( 2 : 1 \)
And \( k = \frac{3m - 4}{m + 1} \)
Put \( m = 2 \)
\( \Rightarrow k = \frac{6 - 4}{3} = \frac{2}{3} \)
Hence, the required ratio is \( 2 : 1 \) and the value of \( k \) is \( \frac{2}{3} \).

Question. A line intersects the y-axis and x-axis at the points \( P \) and \( Q \) respectively. If \( (2, –5) \) is the mid-point of \( PQ \), then find the coordinates of \( P \) and \( Q \). 
Answer: Let coordinates of \( P \) be \( (0, y) \) and of \( Q \) be \( (x, 0) \).
\( M(2, -5) \) is mid point of \( PQ \).
By section formula, \( (2, -5) = \left( \frac{0 + x}{2}, \frac{y + 0}{2} \right) \)
\( 2 = \frac{x}{2} \) and \( -5 = \frac{y}{2} \)
\( \therefore x = 4 \) and \( y = -10 \).
\( \therefore P \) is \( (0, -10) \) and \( Q \) is \( (4, 0) \).

Question. If the distances of \( P(x, y) \) from \( A(5, 1) \) and \( B(–1, 5) \) are equal, then prove that \( 3x = 2y \). 
Answer: \( PA = PB \)
\( \therefore PA^2 = PB^2 \)
By distance formula,
\( (5 - x)^2 + (1 - y)^2 = (-1 - x)^2 + (5 - y)^2 \)
\( \Rightarrow (5 - x)^2 + (1 - y)^2 = (1 + x)^2 + (5 - y)^2 \)
\( 25 - 10x + x^2 + 1 - 2y + y^2 = 1 + 2x + x^2 + 25 - 10y + y^2 \)
\( -10x - 2y = 2x - 10y \)
\( 8y = 12x \)
\( 2y = 3x \) or \( 3x = 2y \).
Hence, proved.

Question. The coordinates of houses of Sonu and Labhoo are \( (7, 3) \) and \( (4, 3) \) respectively. Coordinates of their school is \( (2, 2) \). If both leave their house at the same time in the morning and also reach school in same time, then who travel faster? 
Answer: Distance between Sonu’s house and school \( = \sqrt{(2 - 7)^2 + (2 - 3)^2} \)
[\(\because\) Distance \( = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)]
\( = \sqrt{25 + 1} = \sqrt{26} \)
Distance between Labhoo’s house and school \( z = \sqrt{(2 - 4)^2 + (2 - 3)^2} \)
\( = \sqrt{4 + 1} = \sqrt{5} \)
So, distance of Sonu’s house from school is more. Therefore, Sonu travels faster.

Question. The points \( A(4, 7), B(p, 3) \) and \( C(7, 3) \) are the vertices of a right triangle, right-angled at \( B \). Find the value of \( p \). [
Answer: Here, \( AB^2 + BC^2 = AC^2 \)
\( \Rightarrow (4 - p)^2 + (7 - 3)^2 + (p - 7)^2 + (3 - 3)^2 = (4 - 7)^2 + (7 - 3)^2 \)
\( \Rightarrow (4 - p)^2 + (4)^2 + (p - 7)^2 + 0 = (-3)^2 + (4)^2 \)
\( \Rightarrow 16 + p^2 - 8p + 16 + p^2 - 14p + 49 = 9 + 16 \)
\( \Rightarrow 2p^2 - 22p + 56 = 0 \)
\( \Rightarrow p^2 - 11p + 28 = 0 \)
\( \Rightarrow (p - 7)(p - 4) = 0 \)
\( \Rightarrow p = 7 \) or \( 4 \)
Since, \( p \neq 7 \) (otherwise \( B \) and \( C \) will coincide), \( \therefore p = 4 \).

Question. If the point \( C(– 1, 2) \) divides internally the line segment joining \( A(2, 5) \) and \( B(x, y) \) in the ratio \( 3 : 4 \), find the coordinates of \( B \). 
Answer: Since \( C \) divides \( AB \) in the ratio \( 3 : 4 \), we have
By section formula \( C(x, y) = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right) \)
\( C\left( \frac{3x + 8}{7}, \frac{3y + 20}{7} \right) = (–1, 2) \)
\( \Rightarrow \frac{3x + 8}{7} = -1 \) and \( \frac{3y + 20}{7} = 2 \)
\( \Rightarrow x = -5 \) and \( y = -2 \)
Thus, coordinates of \( B \) are \( (–5, –2) \).

SHORT ANSWER (SA-II) Type Questions

Question. Prove that the points \( (2, – 2), (– 2, 1) \) and \( (5, 2) \) are the vertices of a right angled triangle. Also find the area of this triangle. 
Answer: Given, \( \Delta ABC \), with vertices \( A(2, – 2), B(– 2, 1) \) and \( C(5, 2) \)
To Prove: \( \Delta ABC \) is a right angled triangle.
Proof: We will find the lengths of the sides \( AB, BC \) and \( CA \) by using the distance formula.
Distance formula \( = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
For line \( AB, AB = \sqrt{(-2 - 2)^2 + (1 - (-2))^2} = 5 \) units.
For line \( BC, BC = \sqrt{(5 - (-2))^2 + (2 - 1)^2} = \sqrt{50} \) units.
For line \( CA, CA = \sqrt{(5 - 2)^2 + (2 - (-2))^2} = 5 \) units.
Then, \( BC^2 = 50 \) units and \( AB^2 + AC^2 = 25 + 25 = 50 \) units.
\( \therefore AB^2 + AC^2 = BC^2 \)
By the converse of the pythagoras theorem, the given triangle is right angled at \( A \).
Hence, proved.
Area of \( (\Delta ABC) = \frac{1}{2} \times \text{base} \times \text{height} \)
\( = \frac{1}{2} \times AB \times AC \)
\( = \frac{1}{2} \times 5 \times 5 = 12.5 \) sq. units.

Question. Find the ratio in which y-axis divides the line segment joining the points \( A(5, –6) \) and \( B(–1, –4) \). Also, find the coordinates of the point of division.
Answer: Let, point \( P(0, y) \) be the point that divides the given line-segment. Since, point \( P \) is on y-axis, its x-coordinate is zero.
Let the ratio in which \( P \) divides \( AB \) be \( k : 1 \).
By the section formula \( P(x, y) = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right) \)
\( (0, y) = \left( \frac{k \times (-1) + 1 \times 5}{k + 1}, \frac{k \times (-4) + 1 \times (-6)}{k + 1} \right) \)
On comparing the x-coordinate: \( 0 = \frac{-k + 5}{k + 1} \Rightarrow k = 5 \)
Hence, the required ratio is \( 5 : 1 \).
Now, comparing the y-coordinate: \( y = \frac{5 \times (-4) + 1 \times (-6)}{5 + 1} = \frac{-20 - 6}{6} = \frac{-26}{6} = \frac{-13}{3} \)
Hence, the coordinates of point are: \( P\left( 0, \frac{-13}{3} \right) \).

Question. The x-coordinate of a point \( P \) is twice its y-coordinate. If \( P \) is equidistant from \( Q(2, –5) \) and \( R(–3, 6) \), find the coordinates of \( P \). 
Answer: Let the y-coordinate of \( P \) be ‘\( a \)’. Then, the x-coordinate is ‘\( 2a \)’.
Coordinates of \( P \) are \( (2a, a) \).
Since, point \( P \) is equidistant from \( Q(2, –5) \) and \( R(–3, 6) \), then by the distance formula \( PQ = PR \)
\( \Rightarrow \sqrt{(2a - 2)^2 + (a + 5)^2} = \sqrt{(2a + 3)^2 + (a - 6)^2} \)
On squaring both sides, we get
\( (2a – 2)^2 + (a + 5)^2 = (2a + 3)^2 + (a – 6)^2 \)
\( \Rightarrow 4a^2 - 8a + 4 + a^2 + 10a + 25 = 4a^2 + 12a + 9 + a^2 - 12a + 36 \)
\( \Rightarrow 2a + 29 = 45 \)
\( \Rightarrow 2a = 16 \Rightarrow a = 8 \)
Then, y-coordinate \( = 8 \) and x-coordinate \( = 16 \).
Hence, the coordinates of the required point \( P \) are \( (16, 8) \).

Question. If the distance between the points \( (4, k) \) and \( (1, 0) \) is \( 5 \), what can be the possible values of \( k \)? 
Answer: Given, points are \( A(4, k) \) and \( B(1, 0) \). Distance between points \( A \) and \( B = 5 \).
According to the distance formula: \( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( 5 = \sqrt{(1 - 4)^2 + (0 - k)^2} \)
\( \Rightarrow (5)^2 = (-3)^2 + (-k)^2 \) (on squaring both sides)
\( \Rightarrow 25 = 9 + k^2 \)
\( \Rightarrow k^2 = 25 - 9 = 16 \)
\( \Rightarrow k = \pm\sqrt{16} = \pm 4 \)
Hence, the possible values of \( k \) are \( 4 \) and \( – 4 \).