Coordinate Geometry

1. The distance between the points (2,-3) and (-2, 3) is: (2024)
(a) 2√13 units
(b) 5 units
(c) 13√2 units
(d) 10 units

Answer. (a) 2√13 units

2. The diameter of a circle is of length 6 cm. If one end of the diameter is (-4,
0), the other end on x-axis is at : (2024)
(a) (0,2)
(b) (6,0)
(c) (2,0)
(d) (4,0)

Answer. (c) (2,0)

3.

(2024)

𝟕 𝟗
Answer. (d) ( , )
𝟐 𝟒

4. In the given figure, AP ⊥ AB (2024)

and BQ ⊥AB. If OA = 15 cm,
BO = 12 cm and AP = 10 cm,
then find the length of BQ.
Answer.  OAP   OBQ (AA)

5. The line AB intersects x-axis at A and y-axis at B. The point P(2, 3) lies on
AB such that AP: PB = 3:1. Find the co-ordinates of A and B. (2024)

Answer. Let co-ordinates of point A be (x, 0) and B(0, y)

 x = 8, y = – 4
 Point A is (8, 0) and B is (0, – 4)

6. To keep the lawn green and cool, Sadhna uses water sprinklers which rotate
in circular shape and cover a particular area.
The diagram below shows the circular areas covered by two sprinklers:
(2024)

Two circles touch externally. The sum of their areas is 130 π sq m and the
distance between their centres is 14 m.
Based on above information, answer the following questions:

(i) Obtain a quadratic equation involving R and r from above. (2024)

Answer. R2 + r2 = 130

(ii) Write a quadratic equation involving only r. (2024)

Answer. r2 – 14r + 33 = 0

(iii) (a) Find the radius r and the corresponding area irrigated. (2024)

Answer. (a) r2 – 14r + 33 = 0  (r – 11) (r – 3) = 0

 r = 3 m, r  11 m (As r < R)
Corresponding area irrigated = 9 m2

OR

(b) Find the radius R and the corresponding area irrigated. (2024)

Answer. (b) R2 – 14R + 33 = 0  (R – 11) (R – 3) = 0

 R = 11 m, R  3 (As R>r)
Corresponding area irrigated = 121 m2

7. AD is a median of AABC with vertices A(5, 6), B(6, 4) and C(0, 0). Length AD
is equal to : (2024)
(a) √68 units
(b) 2√15 units
(c) √101 units
(d) 10 units

Answer. (a) √68 units

8. If the distance between the points (3,-5) and (x,-5) is 15 units, then the
values of x are: (2024)
(a) 12,-18
(b) - 12, 18
(c) 18,5
(d) -9,- 12

Answer. (b) – 12, 18

9. The centre of a circle is at (2, 0). If one end of a diameter is at (6, 0), then
the other end is at : (2024)
(a) (0,0)
(b) (4,0)
(c) (-2, 0)
(d) (-6, 0)

Answer. (c) (– 2, 0)

10.

segment joining the points (1, 2) and (2, 3). Also, find the value of y. (2024)

Answer.

OR

(B) ABCD is a rectangle formed by the points A (−1, −1), B (−1, 6), C (3, 6)
and D (3, −1). P, Q, R and S are mid-points of sides AB, BC, CD and DA
respectively. Show that diagonals of the quadrilateral PQRS bisect each other.
(2024)

Answer.
Since coordinates of mid point of QS = coordinates of mid point of PR
Therefore, diagonals PR and QS bisect each other.
7.1 Introduction
MCQ
1. The distance of the point (-1, 7) from x-axis is
(a) -1
(b) 7
(c) 6
(d) √50 (2023)
2. Assertion (A): Point P(0, 2) is the point of intersection
of y-axis with the line 3x + 2y = 4.
Reason (R): The distance of point P(0, 2) from x-axis is 2 units.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct
explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the
correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true. (2023)
3. The line represented by 4x - 3y = 9 intersects the y-axis at

7.2 Distance Formula

MCQ
4. The distance of the point (-6, 8) from origin is
(a) 6
(c) 8
(b) -6
(d) 10 (2023)
5. The points (-4, 0), (4, 0) and (0, 3) are the vertices of a
(a) right triangle
(b) isosceles triangle
(c) equilateral triangle
(d) scalene triangle (2023)
6. The point on x-axis equidistant from the points P(5,0) and Q(-1, 0) is
(a) (2,0)
(b) (-2,0)
(c) (3,0)
(d) (2, 2) (Term I, 2021-22)
7. The x-coordinate of a point P is twice its y-coordinate. If P is equidistant
from Q(2, −5) and R(-3, 6), then the coordinates of P are
(a) (8, 16)
(b) (10,20)
(c) (20, 10)
(d) (16,8) (Term I, 2021-22)
8. If the point (x, 4) lies on a circle whose centre is at the origin and radius is 5
cm, then the value of x is
(a) O
(b) +4
(c) +5
(d) ±3 (Term I, 2021-22)
9. The distance between the points (m, -n) and (-m, n) is
(a) √m²+n²
(b) m+n
(c) 2√m²+n²
(d) √2m²+2n² (2020)
10. The distance between the points (0, 0) and (a – b,a+b) is
(a) 2√ab
(b) √2a2 +ab
(c) 2√a²+b²
(d) √2a2+2b2 (2020 C)
11. 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
(a) 5
(b) 4
(c) 3
(d) 25 (AI 2014)
12. The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(a) 7+√5
(b) 5
(c) 10
(d) 12 (Foreign 2014)
VSA (1 mark)
13. AOBC is a rectangle whose three vertices are A(0, −3), O(0, 0) and B(4, 0).
The length of its diagonal is (2020)
14. Find the value(s) of x, if the distance between the points A(0, 0) and B(x,-
4) is 5 units. (2019) An
15. Find the distance of a point P(x, y) from the origin. (2018) R
16. If the distance between the points (4, k) and (1, 0) is 5, then what can be
the possible values of k? (Delhi 2017)
SAI (2 marks)
17. If the distances of P(x, y) from A(5, 1) and B(-1, 5) are equal, then prove
that 3x = 2y. (AI 2017)
18. 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. (Delhi 2016)
19. Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right
angled isosceles triangle. (AI 2016)
20. 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. (Foreign 2016)
21. If A(5, 2), B(2, -2) and C(-2, t) are the vertices of a right angled triangle
with B = 90°, then find the value of t. (Delhi 2015)
22. 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. (Al 2015)
23. If A(4, 3), B(-1, y) and C(3, 4) are the vertices of a right triangle ABC, right-
angled at A, then find the value of y. (Al 2015)
24. Show that the points (a, a), (-a, -a) and (-√3a,√3a) are the vertices of an
equilateral triangle. (Foreign 2015)
SA II (3 marks)
25. The centre of a circle is (2a, a - 7). Find the values of 'a' if the circle passes
through the point (11, −9). Radius of the circle is 5√2 cm. (2023)
26. Show that the points (7, 10), (-2, 5) and (3,-4) are vertices of an isosceles
right triangle. (2020)
27. Find the point on y-axis which is equidistant from the points (5,-2) and (-3,
2). (Delhi 2019)
28. Show that (a, a), (−a, −a) and (−√3a, √3a) are vertices of an equilateral
triangle. (2019C)
29. Show that AABC, where A(-2, 0), B(2, 0), C(0, 2) and APQR, where P(-4, 0),
Q(4, 0), R(0, 4) are similar triangles. (Delhi 2017)
30. If the point P(x, y) is equidistant from the points A(a + b, b - a) and B(a - b,
a + b). Prove that bx = ay. (AI 2016)
31. If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find
p. Also find the length of AB. (Delhi 2014)
32. 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. (Delhi 2014)
33. If the points P(2, 2) is equidistant from the points A(-2, k) and B(-2k, -3),
find k. Also find the length of AP. (Delhi 2014)
34. If the point P(k − 1, 2) is equidistant from the points A(3, k) and B(k, 5),
find the values of k. (AI 2014)
35. Find a point P on the y-axis which is equidistant from the points A(4, 8)
and B(-6, 6). Also find the distance AP. (AI 2014)
LA (4/5/6 marks)
36. The base BC of an equilateral triangle ABC lies on y-axis. The coordinates
of point C are (0, -3). The origin is the mid-point of the base. Find the
coordinates of the points A and B. Also find the coordinates of another point D
such that BACD is a rhombus. (Foreign 2015)
37. The base QR of an equilateral triangle PQR lies on x-axis. The coordinates
of point Q are (-4, 0) and the origin is the mid-point of the base. Find the
coordinates of the points P and R. (Foreign 2015)
7.3 Section Formula
MCQ
38. In what ratio, does x-axis divide the line segment joining the points A(3, 6)
and B(-12,-3)?
(a) 1:2
(b) 1:4
(c) 4:1
(d) 2:1 (2023)
39. The ratio in which the point (-4, 6) divides the line segment joining the
points A(-6, 10) and B(3, -8) is
(a) 2:5
(b) 7:2
(c) 2:7
(d) 5:2 (Term I, 2021-22)
Case Study: Shivani is an interior decorator. To design her own living room,
she designed wall shelves. The graph of intersecting wall shelves is given
below:
Based on the above information, answer the following questions:
40. If O is the origin, then what are the coordinates of S?
(a) (-6,-4)
(b) (6,4)
(c) (-6,4)
(d) (6,-4) (Term I, 2021-22)
41. The coordinates of the mid-point of the line segment joining D and H is

42. The ratio in which the x-axis divides the line- segment joining the points A
and C is
(a) 2:3
(b) 2:1
(c) 1:2
(d) 1:1 (Term 1, 2021-22)
43. The distance between the points P and G is
(a) 16 units
(b) 3√74 units
(c) 2√74 units
(d) √74 units (Term I, 2021-22)
44. The coordinates of the vertices of rectangle IJKL are
(a) 1(2, 0), J(2, 6), K(8, 6), L(8, 2)
(b) 1(2,-2), J(2, −6), K(8, - 6), L(8,-2)
(c) (-2, 0), J(-2, 6), K(-8, 6), L(-8, 2)
(d) (-2, 0), J(-2, -6), K(-8, -6), L(-8,-2) (Term I, 2021-22)
45. Case Study: Students of a school are standing in rows and columns in their
school playground to celebrate their annual sports day. A, B, C and D are the
positions of four students as shown in the figure.

Based on the above, answer the following questions:

I. The figure formed by the four points A, B, C and D is a
(a) square
(b) parallelogram
(c) rhombus
(d) quadrilateral
II. If the sports teacher is sitting at the origin, then which of the four students
is closest to him?
(a) A
(b) B
(c) C
(d) D
III. The distance between A and C is
(a) √37 units
(b) √35 units
(c) 6 units
(d) 5 units
IV. The coordinates of the mid-point of line segment AC are

V. If a point P divides the line segment AD in the ratio

1:2, then coordinates of P are

46. 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) (2020)
47. 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 kis
(a) 1
(b) 2
(c) -2
(d) -1 (2020)
48. The centre of a circle whose end points of a diameter are (-6, 3) and (6, 4)
is

VSA (1 mark)
49. Find the coordinates of a point A, where AB is a diameter of the circle with
centre (-2, 2) and B is the point with coordinates (3, 4). (Delhi 2019)
50. In what ratio is the line segment joining the points P(3, -6) and Q(5, 3)
divided by x-axis? (2019C)
SAI (2 marks)
51. Find the ratio in which the segment joining the points (1, 3) and (4, 5) is
divided by x-axis? Also find the coordinates of this point on x-axis. (Delhi
2019)
52. The point R divides the line segment AB, where

coordinates of R. (2019)
53. Find the coordinates of a point A, where AB is a diameter of the circle with
centre (3, -1) and the point B is (2, 6). (AI 2019)
54. 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. (2018)
55. 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. (AI 2017)
56. 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. (Delhi
2016)
57. Let P and Q be the points of trisection of the line segment joining the
points A(2,-2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P
and Q. (NCERT, AI 2016)
58. 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. (Foreign 2016)
59.

SA II (3 marks)
60. Find the ratio in which the y-axis divides the line segment joining the
points (6, -4) and (-2, -7). Also, find the point of intersection. (2020)
61. 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. (2020)
62. The line segment joining the points A(2, 1) and B(5, -8) is trisected at the
points P and Q such that P is nearer to A. If P also lies on the line given by 2xy
+ k = 0, find the value of k. (Delhi 2019)
OR
Point P divides the line segment joining the

63. Find the ratio in which the line x - 3y = O divides the line segment joining
the points (-2, -5) and (6, 3). Find the coordinates of the point of intersection.
(2019)
64. In what ratio does the point P(-4, y) divide the line segment joining the
points A(-6, 10) and B(3, -8)? Hence find the value of y. (2019)
65. If A(-2, 1), B(a, O), C(4, b) and D(1, 2) are the vertices of a parallelogram
ABCD, find the values of a and b. Hence find the lengths of its sides. (2018)
66.

67. If the point C(-1, 2) divides internally the line-segment joining the points
A(2, 5) and B(x, y) in the ratio 3: 4, find the value of x² + y². (Foreign 2016)
68. If the coordinates of points A and B are (−2, −2) and (2, -4) respectively,
find the coordinates

69. Find the coordinates of a point P on the line segment

70. Point A lies on the line segment PQ joining P(6,-6)

also lies on the line 3x + k(y + 1) = 0, find the value of k. (Foreign 2015)
71. 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 coordinates of the point of division. (AI
2014)
72. 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. (AI 2014)
73. Points P, Q, R and S divide the line segment joining the points A(1, 2) and
B(6, 7) in 5 equal parts. Find the coordinates of the points P, Q and R. (Foreign
2014)
LA (4/5/6 marks)
74. Case Study: Jagdish has a field which is in the shape of a right angled
triangle AQC. He wants to leave a space in the form of a square PQRS inside the
field for growing wheat and the remaining for growing vegetables (as shown
in the figure). In the field, there is a pole marked as O.

Based on the above information, answer the following questions:

(i) Taking O as origin, coordinates of P are (-200, 0) and of Q are (200, 0).
PQRS being a square, what are the coordinates of R and S?
(ii) (a) What is the area of square PQRS?
OR
(b) What is the length of diagonal PR in square PQRS?
(iii) If S divides CA in the ratio K: 1, what is the value of K, where point A is
(200, 800) ? (2023)
75. Find the ratio in which the point P(x, 2) divides the line segment joining
the points A(12, 5) and B(4, -3). Also find the value of x.
(Delhi 2014)
76. The mid-point 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. (Foreign 2014)

CBSE Sample Questions

7.2 Distance Formula
MCQ
1. In the question, a statement of Assertion (A) is followed by a statement of
Reason (R). Choose the correct option.
Statement A (Assertion): If the co-ordinates of the mid-points of the sides AB
and AC of AABC are D(3, 5) and E(-3, -3) respectively, then BC = 20 units.
Statement R (Reason): The line joining the mid points of two sides of a
triangle is parallel to the third side and equal to half of it.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true and reason (R) is not the
correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.(2022-23)
2. The distance of the point A(-5, 6) from the origin is
(a) 11 units
(b) 61 units
(c) √11 units
(d) √61 units (Term I, 2021-22)
3. The equation of the perpendicular bisector of line segment joining points
A(4, 5) and B(-2, 3) is
(a) 2x - y +7=0
(b) 3x+2y-7=0
(c) 3x-y-7=0
(d) 3x + y -7=0 (Term I, 2021-22)
SAI (2 marks)
4. Find the point on x-axis which is equidistant from the points (2,-2) and (-4,
2). (2020-21)
7.3 Section Formula
MCQ
5. If the vertices of a parallelogram PQRS taken in order are P(3, 4), Q(−2, 3)
and R(-3, -2), then the coordinates of its fourth vertex S are
(a) (-2, -1)
(b) (-2,-3)
(c) (2,-1)
(d) (1, 2) (2022-23)
6. The vertices of a parallelogram in order are A(1, 2), B(4, y), C(x, 6) and D
(3, 5). Then (x, y) is
(a) (6,3)
(b) (3,6)
(c) (5,6)
(d) (1,4) (Term I, 2021-22)
7. Point P divides the line segment joining R(-1, 3) and S(9, 8) in ratio k: 1. If P
lies on the line x − y + 2 = 0, then value of k is

Case Study: A hockey field is the playing surface for the game of hockey.
Historically, the game was played on natural turf (grass) but nowadays it is
predominantly played on an artificial turf.
It is rectangular in shape - 100 yards by 60 yards. Goals consist of two upright
posts placed equidistant from the centre of the backline, joined at the top by a
horizontal cross-bar. The inner edges of the posts must be 3.66 metres (4
yards) apart, and the lower edge of the crossbar must be 2.14 metres (7 feet)
above the ground. Each team plays with 11 players on the field during the
game including the goalie. Positions you might play include-
• Forward: As shown by players A, B, C and D.
• Midfielders: As shown by players E, F and G
• Fullbacks: As shown by players H, I and J
• Goalie: As shown by player K
Using the picture of hockey field below, answer the questions that follow:

8. The coordinates of the centroid of AEHJ are

9. If a player P needs to be at equal distances from A and G, such that A, P and

G are in straight line, then position of P will be given by

10. The point on x axis equidistant from I and E is

11. What are the coordinates of the position of a player Q such that his
distance from K is twice his distance from E and K, Q and E are collinear?
(a) (1, 0)
(b) (0, 1)
(c) (-2, 1)
(d) (-1, 0) (Term I, 2021-22)
12. The point on y-axis equidistant from B and C is
(a) (-1, 0)
(b) (0, -1)
(c) (1,0)
(d) (0, 1) (Term I, 2021-22)
Case study-based questions are compulsory. Attempt any 4 sub parts from
given question. Each question carries 1 mark.
13. Sun Room: The diagrams show the plans for a sun room. It will be built
onto the wall of a house. The four walls of the sunroom are square clear glass
panels. The roof is made using
• Four clear glass panels, trapezium in shape, all the same size
• One tinted glass panel, half a regular octagon in shape.

(i) Refer to Top View

Find the mid-point of the segment joining the points J(6, 17) and (9, 16).
(a) (33/2, 15/2)
(b) (3/2, 1/2)
(c) (15/2, 33/2)
(d) (1/2, 3/2)
(ii) Refer to Top View
The distance of the point P from the y-axis is
(a) 4
(b) 15
(iii) Refer to Front View
(c) 19
(d) 25
The distance between the points A and S is
(a) 4
(b) 8
(c) 16
(d) 20
(iv) Refer to Front View
Find the co-ordinates of the point which divides the line segment joining the
points A and B in the ratio
1:3 internally.
(a) (8.5, 2.0)
(b) (2.0, 9.5)
(c) (3.0, 7.5)
(d) (2.0, 8.5)
(v) Refer to Front View
If a point (x, y) is equidistant from the Q(9, 8) and S(17,8), then
(a) x + y = 13
(b) x-13=0
(c) y-13=0
(d) x - y = 13 (2020-21)
SAI (2 marks)
14. P (-2, 5) and Q (3, 2) are two points. Find the co-ordinates of the point R
on PQ such that PR = 2QR. (2020-21)
LA (4/5/6 marks)
Case study-based questions are compulsory. Attempt any 4 sub parts from
given question.
15. A tiling or tessellation of a flat surface is the covering of a plane using one
or more geometric shapes, called tiles, with no overlaps and no gaps.
Historically, tessellations were used in ancient Rome and in Islamic art. You
may find tessellation patterns on floors, walls, paintings etc. Shown below is a
tiled floor in the archaeological Museum of Seville, made using squares,
triangles and hexagons.

A craftsman thought of making a floor pattern after being inspired by the

above design. To ensure accuracy in his work, he made the pattern on the
Cartesian plane. He used regular octagons, squares and triangles for his floor
tessellation pattern.

Use the above figure to answer the questions that follow:

(i) What is the length of the line segment joining points B and F?
(ii) The centre 'Z' of the figure will be the point of intersection of the diagonals
of quadrilateral WXOP. Then what are the coordinates of Z?
(iii) What are the coordinates of the point on y-axis equidistant from A and G?
OR
What is the area of Trapezium AFGH? (2022-23)
 SOLUTIONS
Previous Years' CBSE Board Questions
1. (b): Distance from x-axis = y-coordinate of point = 7 units
2. (b): Point P(0, 2) is the point of intersection of y-axis with line 3x + 2y = 4.

Also, the distance of point P(0, 2) from x-axis is 2 units.

:- Both assertion and Reason are true but Reason is not the correct
explanation of Assertion.
3. (a): Given, the equation of line is 4x-3y = 9.
Putting x = 0, we get 4x0-3y=9 ⇒ y=-3
So, the line 4x-3y=9 intersects the y-axis at (0, -3).
4. (d): Distance of the point (-6, 8) from origin (0, 0)
=√(−6-0)²+(8-0)² =√36+64 = √100=10 units
5. (b): The points be A(-4, 0), B(4, 0) and C(0, 3).
Using distance formula d=√(x₂−x₁)²+(y₂−Y₁)²

And, AB² + BC² + CA2

BC= CA
:- ∆AABC is an isosceles triangle.
6. (a): Let coordinates of the point on the x-axis be
R (x, 0).
Given, PR = QR ⇒ PR² = QR²
= (x-5)² + (0 - 0)² = (x + 1)² + (0-0)²
= x² - 10x+25= x² + 2x + 1⇒ 12x = 24⇒ x=2
Required point is (2, 0).
7. (d): Let y-coordinate of point P = t
So, x-coordinate of point P = 2t . Point is P (2t, t).
Given, PQ = RP⇒ PQ² = RP2
⇒ (2t-2)²+(t+5)² = (2t + 3)² + (t − 6)² [By distance formula]
⇒ 4t² - 8t+4+t2+10t+25= 4t² + 12t+9+t²-12t+36
⇒2t = 16 ⇒ t = 8
:- Coordinates of P are (16, 8).
8. (d): Given, (x, 4) lies on the circle and coordinates of
centre is origin i.e. (0, 0)
So, radius = √(x-0)² +(4–0)²
=√(x−0)²+(4-0)2
⇒ 5=√x²+1625 = x²+16 (Given, radius = 5 cm)
⇒ x²=9⇒x=±3
9. (c): Required distance

10. (d): Required distance = √(a-b-0)²+(a+b-0)²

11. (a) ABCD is a rectangle having vertices B(4, 0),

C(4,3) and D(0, 3). Using distance formula,

12. (d): Let O(0, 0), A(3, 0) and B(0, 4) represent the ∆ABO.
:- OA = 3 units, OB = 4 units and

:- Perimeter of ∆OAB = OA+OB+ BA

=3+4+5=12 units
13. In rectangle AOBC, AB is a diagonal.

14. Given AB = 5 units

⇒ x² + (4)² = (5)² ⇒ x² + 16 = 25 ⇒ x²=9⇒x=±3

15. Distance of point P(x, y) from the origin O(0, 0) is
given by

16. Let A(4, k) and B(1, 0).

Also, AB = 5 (Given)

⇒ 9+k² = 25⇒ k² = 16 ⇒ k = ±4
17. We have, P(x, y), A(5, 1) and B(-1, 5)
Given, PA = PB

= x²+25-10x+y²+1-2y= x²+1+2x+y²+25-10y
= -10x - 2y=2x-10y =>12x = -8y => 3x = 2y
18. Let the required point be P(2y, y).
Since P(2y, y) is equidistant from given points Q(2,-5) and
R(-3, 6).
:- PQ = PR ⇒ PQ² = PR²
⇒ (2y-2)² + (y + 5)² = (−3 − 2y)² + (6 − y)²
⇒ 4y²+4-8y+y²+25+10y=9+4y²+12y+36+ y²-12y
⇒ 2y= 16 ⇒ y=8
Hence, the required point P is (16,8).
19. Let A(3, 0), B(6, 4) and C(−1, 3) be the vertices of triangle.

(By using distance formula)

AC=√(3+1)²+(0-3)² = √16+9=5units
= √(6+1)² +(4−3)² = √50=5√2 units
Now, as AB = AC and (AB)2 + (AC)² = (BC)2
:- AABC is a right angled isosceles triangle.
20. Let A(2,-2), B(-2, 1) and C(5, 2)

Here, (AB)² + (AC)² = 25+ 25 = 50 = (BC)²

Hence, AABC is a right angled triangle at A.

21. By using Pythagoras theorem in right angled AABC, we have

AC² = AB² + BC2
⇒ (-2-5)²+(t-2)² = (2-5)²+(−2−2)²+(-2-2)²+(t+2)²
⇒ 49+t²-4t+ 4 = 9 + 16 + 16 + t² + 4t+4
⇒ -8t = -8 ⇒ t = 1
22. By Pythagoras theorem in AABC, we have AB²+BC2=AC2
⇒ (p −4)² + (3 −7)² + (7 − p)² + (3 − 3)² = (7-4)² + (3-7)²
= p² - 8p + 16 + 16 + 49 - 14p+ p²+0=9+16
⇒ 2p² - 22p+ 56 = 0⇒ p² - 11p + 28 = 0
⇒ (p-7)(p-4)=0⇒p=7 or p=4

Hence, p = 4
23. By Pythagoras theorem in AABC, we have
AB² + AC² = BC2
⇒ (-1-4)²+(y-3)² + (3 − 4)² + (4 − 3)² = (3 + 1)² + (4 − y)²
⇒ 25+ y² - 6y+ 9 + 1 + 1 = 16 + 16 - 8y + y²
⇒ -6y+8y = 32-36 2y=-4⇒ y = -2
24. Let the vertices be A(a, a), B(-a, -a) and C(-√3a, √3a).

Hence, AB = BC = CA
Therefore, ABC is an equilateral triangle.
25. Given centre of a circle is (2a, a - 7)

Radius of circle = 5√2 cm

:- Distance between centre (2a, a - 7) and (11, -9)
= radius of circle.

⇒ (11-2a)²+(-2-a)² = 25x2=50
⇒ 121+ 4a² - 44a+4+ a²+4a= 50
⇒ a² - 8a +15=0⇒ (a − 3) (a − 5) = 0⇒ a = 3 or a = 5
26. Let the given points be A(7, 10), B(-2, 5) and C(3,-4).

Since, AB BC. ABC is an isosceles triangle.

Also, AB² + BC² = 106 + 106 = 212 = AC2
So, ABC is an isosceles right angled triangle with /B = 90°.
27. Let P(O, y) be the point on the y-axis which is
equidistant from A(5, 2) and B(-3, 2).
:- AP = BP⇒ (AP)² = (BP)²
⇒ (5-0)²+(-2 − y)² = (-3 − 0)² + (2 − y)²
⇒ 25+ 4+ y²+ 4y = 9+4+ y²-4y

Hence, A and B are equidistant from (0, -2).

28. Let A(a, a), B(-a, -a), C(−√3a,√3a) be vertices of a triangle.

Hence, A(a, a), B(-a, -a) and C(-√3a, √3a) are vertices of an equilateral
triangle.
29. By using distance formula,

30. Given, A(a + b, b − a) and B(a−b, a + b) are equidistant

from P(x, y).
i.e., PA = PB (PA)² = (PB)2
⇒ [x-(a+b)]² + [y − (b − a)]² = [(x − (a - b)]² + [y − (a + b)]²
⇒ (a+b)² - 2(a + b)x + (b - a)² - 2(b - a)y
= (a - b)² - 2(a - b)x + (a + b)² −2(a + b)y
⇒ 2[(a + b)x + (b - a)y] = 2[(a - b)x + (a + b)y]
⇒ (a + b)x + (b - a)y = (a - b)x + (a + b)y
⇒ (a + b = a + b)x = (a + b − b + a)y
⇒ 2bx = 2ay or bx = ay.
31. The given points are A(0, 2), B(3, p) and C(p, 5). Since A is equidistant
from B and C
:- AB = AC => AB² = AC2
⇒ (3-0)² + (p-2)² = (p-0)² + (5-2)2
⇒ 9+ p² + 4-4p = p²+9=4-4p = 0
⇒ 4p = 4⇒ p=1
32. Points A(-1, y) and B(5, 7) lie on a circle with centre O(2, -3y)

33. Given, P(2, 2), A(-2, k) and B(-2k, -3)

Since P is equidistant from A and B.
.. AP BP AP² = BP2
⇒ (2 + 2)² + (2 −k)² = (2 + 2k)² + (2 + 3)²
⇒ 16+4+k² - 4k = 4 + 4k² + 8k + 25
⇒ 3k² + 12k + 9 = 0 ⇒ k² + 4k +3=0
⇒ k2 + 3k + k + 3 = 0 ⇒ k(k + 3) + 1(k + 3) = 0
⇒ (k + 1)(k + 3) = 0 ⇒ k = -1 or k = -3

34. Since P is equidistant from points A and B

:- AP = BP⇒ AP2 = BP2
⇒ (k-1-3)² + (2 − k)² = (k−(k − 1))² + (5- 2)2
⇒ (k − 4)² + (2 −k)² = (1)² + (3)²
⇒ k² + 16 - 8k + 4+k² - 4k = 10
⇒ 2k² - 12k +20= 10 ⇒ 2k² - 12k + 10 = 0
⇒ k² - 6k+5=0⇒ (k − 1)(k − 5) = 0 ⇒ k = 1,5
35. Let P be (0, y) which is equidistant from A(4, 8) and B(-6, 6).

36. :- O is the mid-point

of the base BC,
:- Coordinates of point B
are (0,3)
So, BC = 6 units.
Let the coordinates of
point A be (x, 0).
37. The coordinates of Q are (-4,0).
Since, O(0, 0) is the mid-point of QR.
:- Coordinates of point R are (4, 0).
Let coordinates of point P be (0, y) or (0, -y)

Now, PQ = QR
⇒ PQ² = QR2
⇒ (A is equilateral)
⇒ (-4-0)²+(0-y)² = (4 + 4)² + (0-0)²
⇒ 16+ y² = 64⇒ y² = 64-16 ⇒ y² =
⇒ y=+4√3
:- The coordinates of point P are (0,4√3) or (0,-4√3).
38. (d): Let the point on the x-axis be (x, 0), which divides the line segment
joining the points A(3, 6) and B(-12, -3) in the ratio k : 1.

Hence, the required ratio is 2: 1.

39. (c): Let point P(-4, 6) divides the line segment AB in the ratio m₁: m2.

40. (c): Coordinates of S are (-6, 4).

41. (b): Coordinates of D are (-2, -4) and coordinates of H are (8, 2).

42. (d): Coordinates of A are (-2, 4) and coordinates of C are (4, - 4).
Let (x, 0) divides the line segment joining the points A and
C in the ratio m₁: m2.
By section formula, we have

43. (c): Coordinates of P are (-6, -4) and coordinates of Gare (8,6).
44. (b): Coordinates of vertices of rectangle IJKL are respectively |(2, −2), J(2,
−6), K(8, -6), L(8,-2).
45. I. (d): From figure coordinates are A(2, 5), B(5, 7), C(8,6) and D(6, 3).

Clearly, ABCD is a quadrilateral

II. (a): Here, sports teacher is at O(0, 0).

:- OA is the minimum distance

:- A is closest to sports teacher.
V. (b): Let point P(x, y) divides the line segment AD in the ration 1: 2.

46. (d): Let coordinates of the point on the x-axis be P(x, 0). Let the given
points be A(-4, 0) and B(10, 0), which also lie on x-axis.
Since P is equidistant from A and B.

So, P(3, 0) is equidistant from A(-4, 0) and B(10, 0).

47. (d) Since, the point P(k, O) divides the line segment joining A(2,-2) and B
(-7, 4) in the ratio 1 : 2.

48. (c): Let the coordinates of centre of the circle be (x, y) and AB be the given
diameter.
:- By using mid-point formula,

49. Let coordinates of the point A be (x, y) and O is the mid point of AB. By
using mid-point formula, we have
50. Let the point R(x, 0) on x-axis divides the line segment PQ in the ration k:
1.

51. Let the point P(x, 0) divides the segment joining the points A (1,-3) and B
(4, 5) in the ratio k : 1.

53. Let the coordinates of A be (x, y).

Here, O(3, 1) is the mid point of AB.
:- By using mid point formula, we have
54. The given points are A(2, 3) and B(6, -3). Let the point P(4, m) divides the
line segment AB in the ratio k : 1.

55. Let coordinates of P and Q be (0, y) and (x, 0) respectively.

Let M(2,5) be the mid-point of PQ.

56. Let the point P(O, y) on y-axis divides the line segment AB in the ratio k : 1.
57. Let (X1, Y₁) and (x2, y2) be coordinates of P and Q respectively.

58. Let the point (-3, k) divides the line segment joining the points (-5,-4) and
(-2, 3) in the ratio λ : 1.
59.

60. Let the point P(O, y) on y-axis divides the line segment joining the points
A(6, - 4) and B(-2, -7) in the ratio k: 1.

61. We have, A(2, 5), B(x, y) and C(-1, 2) and point C divides AB in the ratio 3:
4.
62.

Let P(X1, Y₁) and Q(x2, y2) are the points of trisection of line segment AB.
:- AP = PQ = QB
Now, point P divides AB internally in the ratio 1: 2.
:- By section formula, we have

63. Let point P(x1, Y1) divides the line segment joining the points A(-2, -5)
and B(6, 3) in the ratio k: 1.

64. Let the point P(-4, y) divides the line segment joining the points A and B in
the ration k: 1.
65. Let the given points are A(-2, 1), B(a, 0), C(4, b) and D(1, 2)
66.

67. We have, A(2, 5), B(x, y) and C(-1, 2) and point C divides AB in the ratio 3:
4.
68.

69.
70. Point P(6, -6) lies on the line 3x + k(y + 1) = 0

71. Let point P(x, 0) divides the line segment joining the points A and B in the
ratio k: 1.

72. ABCD is a rectangle. AC and

BD are its diagonals.
To prove : AC = BD
and AC bisects BD or vice versa.
73. Since, P, Q, R and S divides the line segment joining the points A(1, 2) and
B(6, 7) in 5 equal parts.

:- P divides AB in ratio 1 : 4.
Let coordinates of P be (x, y).
Using section formula, coordinate of P are given by

74. (i) We have, P= (-200, 0) and Q = (200, 0)

The coordinates of R and S are (200, 400) and (-200, 400).
(ii) (a) The length PQ = 200+ 200
= 400 units.
Area of square PQRS = 400 x 400
= 160000 sq. units.
OR
(b) Length of diagonal PR = √2 × length of side
=400√√2 units.
75. Let the point P(x, 2) divides the line segment joining the points A(12, 5)
and B(4, −3) in the ratio k : 1.

Thus, the value of x is 9.

Also, the point P divides the line segment joining the
points A(12, 5) and B(4, -3) in the ratio 3: 5.
76

CBSE Sample Questions

1.

We know, line joining the mid points of two sides of a triangle is parallel to
third side and equal to half of it.

We know, line joining the mid points of two sides of a triangle is parallel to
third side and equal to half of it.
2. (d): Distance of point A(-5, 6) from the origin
3. (d): Any point P(x, y) of perpendicular bisector will be equidistant from A
and B.

= x²+16 - 8x + y²+25 - 10y = x² + 4 + 4x + y² + 9 − 6y

= -12x-4y+ 28-03x+y-7=0 (1)
4. Let P(x, 0) be the point on x-axis which is equidistant
from A(2,-2) and B(-4, 2).
Thus, PA = PB (1/2)

5. (c): Let coordinates of S is (h, k).

Diagonals of parallelogram bisects each other.

6. (a): Since, ABCD is a parallelogram, diagonals AC and

BD bisect each other.
:- Mid point of AC = Mid point of BD
7. (a): Since, P divides the line segment joining R(-1,3) and S(9,8) in the ratio
k: 1.

8. (a): We have, coordinates of E, H and J are (2, 1),

(-2, 4) and (-2,-2) respectively.
:- Centroid of AEHJ

9. (c): If P needs to be at equal distance from A(3, 6) and G(1,-3) such that A, P
and G are collinear, then P will be the midpoint of AG.

10. (a): Let the point on x-axis which is equidistant from (-1, 1) and E(2, 1) be
(x, 0).

11. (b): Let the coordinates of the position of a player Q such that his distance
from K(-4, 1) is twice his distance from E(2, 1) be Q(x, y).

12. (d): Let the point on y-axis which is equidistant from B(4,3) and C(4, -1)
be (0, y).

13. (i) (c): Mid point of the line segment joining the points

(ii) (a): The distance of the point P from y-axis is 4 units.

(iii) (c) Coordinates of A and S are (1, 8) and (17, 8) respectively, therefore the
distance between the points

(iv) (d): Let M(x, y) divides line segment joining A(1, 8) and B(5, 10) in the
ratio 1 : 3.

(v) (b): Let L(x, y) is equidistant from Q(9, 8) and S(17, 8).
Then, LQLS = (LQ)² = (LS)²

14. Let the coordinates of R be (x, y), clearly R divides the line segment joining
P(-2, 5) and Q(3, 2) in the ratio
2:1. [ PR = 2QR (Given)] (1/2)
15. (i) From the given figure, B(1, 2), F(−2, 9) The distance between the
points B and F is;

(ii) W(-6, 2), X(-4, 0), O(5, 9), P(3, 11)

Clearly, WXOP is a rectangle.
Point of intersection of diagonals of a rectangle is the mid point of the
diagonals. So the required point is mid point of WO or XP

(iii) A(-2, 2), G(-4, 7)

Let the point on y-axis be Z(0, y).
Now, point Z is equidistant from A and G.
:- AZ2 = GZ² (1/2)
⇒ (0+2)²+(y-2)² = (0+ 4)2 + (y - 7)²
⇒ (2)² + y²+4-4y = (4)2 + y²+49-14y (1/2)
⇒ 8-4y=65-14y = 10y = 57
So, y = 5.7
:- The required point is (0, 5.7) (1)

From the given figure,

A(-2, 2), F(-2, 9), G(-4, 7), H(-4, 4) (1/2)
Clearly GH = 7-4 = 3 units
AF=9-27 units
So, height of the trapezium AFGH = 2 units (1/2)