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Class 10 (SSC Board)Mathematics & Statistics2026-27 Syllabus

Chapter 5 Co-ordinate Geometry Problem Set 5 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 5 Co-ordinate Geometry Problem Set 5. Step-by-step solved exercises, numerical problems, and digest answers.

19 Solved Questions• 2845 words

Problem Set 5 Geometry 10th Std Maths Part 2 Answers Chapter 5 Co-ordinate Geometry

Question 1 Maharashtra Board Solution
Fill in the blanks using correct alternatives.

i. Seg AB is parallel to Y-axis and co-ordinates of point A are (1, 3), then co-ordinates of point B can be _______.
(A) (3,1)
(B) (5,3)
(C) (3,0)
(D) (1,-3)

Solution & Step-by-Step Answer:
(D)
Since, seg AB || Y-axis.
∴ x co-ordinate of all points on seg AB
will be the same,
x co-ordinate of A (1, 3) = 1
x co-ordinate of B (1, – 3) = 1
∴ Option (D) is correct.

ii. Out of the following, point lies to the right of the origin on X-axis.
(A) (-2,0)
(B) (0,2)
(C) (2,3)
(D) (2,0)
Answer: (D)

iii. Distance of point (-3, 4) from the origin is _________.
(A) 7
(B) 1
(C) 5
(D) -5
Answer: (C)
Distance of (-3, 4) from origin

iv. A line makes an angle of 30° with the positive direction of X-axis. So the slope of the line is ________.
(A)
(B)
(C)
(D)
Answer: (C)

Question 2 Maharashtra Board Solution
Determine whether the given points are collinear. i. A (0, 2), B (1, -0.5), C (2, -3) ii. P(1,2), Q(2,),R(3,) iii L (1, 2), M (5, 3), N (8, 6)
Solution & Step-by-Step Answer:
∴ slope of line AB = slope of line BC ∴ line AB || line BC Also, point B is common to both the lines. ∴ Both lines are the same. ∴ Points A, B and C are collinear.


∴ slope of line PQ = slope of line QR
∴ line PQ || line QR
Also, point Q is common to both the lines.
∴ Both lines are the same.
∴ Points P, Q and R are collinear.


∴ slope of line LM ≠ slope of line MN
∴ Points L, M and N are not collinear.
[Note: Students can solve the above problems by using distance formula.]

Question 3 Maharashtra Board Solution
Find the co-ordinates of the midpoint of the line segment joining P (0,6) and Q (12,20).
Solution & Step-by-Step Answer:
P(x1,y1) = P (0, 6), Q(x2, y2) = Q (12, 20) Here, x1 = 0, y1 = 6, x2 = 12, y2 = 20 ∴ Co-ordinates of the midpoint of seg PQ ∴ The co-ordinates of the midpoint of seg PQ are (6,13).
Question 4 Maharashtra Board Solution
Find the ratio in which the line segment joining the points A (3, 8) and B (-9, 3) is divided by the Y-axis.
Solution & Step-by-Step Answer:
Let C be a point on Y-axis which divides seg AB in the ratio m : n. Point C lies on the Y-axis ∴ its x co-ordinate is 0. Let C = (0, y) Here A (x1,y1) = A(3, 8) B (x2, y2) = B (-9, 3) ∴ By section formula, ∴ Y-axis divides the seg AB in the ratio 1 : 3.
Question 5 Maharashtra Board Solution
Find the point on X-axis which is equidistant from P (2, -5) and Q (-2,9).
Solution & Step-by-Step Answer:
Let point R be on the X-axis which is equidistant from points P and Q. Point R lies on X-axis. ∴ its y co-ordinate is 0. Let R = (x, 0) R is equidistant from points P and Q. ∴ PR = QR ∴ (x – 2)2 + [0 – (-5)]2 = [x – (- 2)]2 + (0 – 9)2 …[Squaring both sides] ∴ (x – 2)2 + (5)2 = (x + 2)2 + (-9)2 ∴ 4 – 4x + x2 + 25 = 4 + 4x + x2 + 81 ∴ – 8x = 56 ∴ x = -7 ∴ The point on X-axis which is equidistant from points P and Q is (-7,0).
Question 6 Maharashtra Board Solution
Find the distances between the following points. i. A (a, 0), B (0, a) ii. P (-6, -3), Q (-1, 9) iii. R (-3a, a), S (a, -2a)
Solution & Step-by-Step Answer:
i. Let A (x1, y1) and B (x2, y2) be the given points. ∴ x1 = a, y1 = 0, x2 = 0, y2 = a By distance formula, ∴ d(A, B) = a units

ii. Let P (x1, y1) and Q (x2, y2) be the given points.
∴ x1= -6, y1= -3, x2= -1, y2= 9
By distance formula,

∴ d(P, Q) = 13 units

iii. Let R (x1, y1) and S (x2, y2) be the given points.
∴ x1= -3a, y1= a, x2= a, y2= -2a
By distance formula,

∴ d(R, S) = 5a units

Question 7 Maharashtra Board Solution
Find the co-ordinates of the circumcentre of a triangle whose vertices are (-3,1), (0, -2) and (1,3).
Solution & Step-by-Step Answer:
Let A (-3, 1), B (0, -2) and C (1, 3) be the vertices of the triangle. Suppose O (h, k) is the circumcentre of ∆ABC. ∴ (h + 3)2 + (k – 1)2 = h2 + (k + 2)2 ∴ h2 + 6h + 9 + k2 – 2k + 1 = h2 + k2 + 4k + 4 ∴ 6h – 2k + 10 = 4k + 4 ∴ 6h – 2k – 4k = 4 – 10 ∴ 6h – 6k = – 6 ∴ h – k = -1,..(i)[Dividing both sides by 6] OB = OC …[Radii of the same circle] ∴ h2 + (k + 2)2 = (h – 1)2 + (k – 3)2 ∴ h2 + k2 + 4k + 4 = h2 – 2h + 1 + k2 – 6k + 9 ∴ 4k + 4 = -2h + 1 – 6k + 9 ∴ 2h+ 10k = 6 ∴ h + 5k = 3 …(ii) Subtracting equation (ii) from (i), we get ∴ The co-ordinates of the circumcentre of the triangle are (,)
Question 8 Maharashtra Board Solution
In the following examples, can the segment joining the given points form a triangle? If triangle is formed, state the type of the triangle considering sides of the triangle. i. L (6, 4), M (-5, -3), N (-6, 8) ii. P (-2, -6), Q (-4, -2), R (-5, 0) iii. A(,),B(-,-),C(,)
Solution & Step-by-Step Answer:
i. By distance formula, ∴ d(M, N) + d (L, N) > d (L, M) ∴ Points L, M, N are non collinear points. We can construct a triangle through 3 non collinear points. ∴ The segment joining the given points form a triangle. Since MN ≠ LN ≠ LM ∴ ∆LMN is a scalene triangle. ∴ The segments joining the points L, M and N will form a scalene triangle.

ii. By distance formula,


∴ d(P, Q) + d(Q, R) = d (P, R) …[From (iii)]
∴ Points P, Q, R are collinear points.
We cannot construct a triangle through 3 collinear points.
∴ The segments joining the points P, Q and R will not form a triangle.

iii. By distance formula,

∴ d(A, B) + d(B, C) + d(A, C) … [From (iii)]
∴ Points A, B, C are non collinear points.
We can construct a triangle through 3 non collinear points.
∴ The segment joining the given points form a triangle.
Since, AB = BC = AC
∴ ∆ABC is an equilateral triangle.
∴ The segments joining the points A, B and C will form an equilateral triangle.

Question 9 Maharashtra Board Solution
Find k, if the line passing through points P (-12, -3) and Q (4, k) has slope .
Solution & Step-by-Step Answer:
P(x1,y1) = P(-12,-3), Q(X2,T2) = Q(4, k) Here, x1 = -12, x2 = 4, y1 = -3, y2 = k But, slope of line PQ (m) is ….[Given] ∴ = ∴ = k + 3 ∴ 8 = k + 3 ∴ k = 5 The value of k is 5.
Question 10 Maharashtra Board Solution
Show that the line joining the points A (4,8) and B (5, 5) is parallel to the line joining the points C (2, 4) and D (1,7). Proof: ∴ Slope of line AB = Slope of line CD Parallel lines have equal slope. ∴ line AB || line CD

Question 11 Maharashtra Board Solution
Show that points P (1, -2), Q (5, 2), R (3, -1), S (-1, -5) are the vertices of a parallelogram. Proof: By distance formula, In ꠸PQRS, PQ = RS … [From (i) and (iii)] QR = PS … [From (ii) and (iv)] ∴ ꠸ PQRS is a parallelogram. [A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent] ∴ Points P, Q, R and S are the vertices of a parallelogram.

Question 12. Show that the ꠸PQRS formed by P (2, 1), Q (-1, 3), R (-5, -3) and S (-2, -5) is a rectangle. Proof: By distance formula, In ꠸PQRS, PQ = RS …[From (i) and (iii)] QR = PS …[From (ii) and (iv)] ꠸PQRS is a parallelogram. [A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent] In parallelogram PQRS, PR = QS … [From (v) and (vi)] ∴ ꠸PQRS is a rectangle. [A parallelogram is a rectangle if its diagonals are equal]

Question 13. Find the lengths of the medians of a triangle whose vertices are A (-1, 1), B (5, -3) and C (3,5).

Solution & Step-by-Step Answer:
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Solution & Step-by-Step Answer:
Suppose AD, BE and CF are the medians. ∴ Points D, E and F are the midpoints of sides BC, AC and AB respectively. ∴ By midpoint formula, ∴ The lengths of the medians of the triangle 5 units, 2 units and units.
Question 14 Maharashtra Board Solution
Find the co-ordinates of centroid of the triangle if points D (-7, 6), E (8, 5) and F (2, -2) are the mid points of the sides of that triangle.
Solution & Step-by-Step Answer:
Suppose A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of the triangle. D (-7, 6), E (8, 5) and F (2, -2) are the midpoints of sides BC, AC and AB respectively. Let G be the centroid of ∆ABC. D is the midpoint of seg BC. By midpoint formula, E is the midpoint of seg AC. By midpoint formula, Adding (i), (iii) and (v), x2 + x3 + x1 + x3 + x1 + x2 = -14 + 16 + 4 ∴ 2x1 + 2x2 + 2x3 = 6 ∴ x1 + x2 + x3 = 3 …(vii) Adding (ii), (iv) and (vi), y2 + y3 + y1 + y3 + y1 +y2 = 12 + 10 – 4 ∴ 2y1 + 2y2 + 2y3 = 18 ∴ y1 + y2 + y3 = 9 …(viii) G is the centroid of ∆ABC. By centroid formula, ∴ The co-ordinates of the centroid of the triangle are (1,3).
Question 15 Maharashtra Board Solution
Show that A (4, -1), B (6, 0), C (7, -2) and D (5, -3) are vertices of a square. Proof: By distance formula, ∴ □ABCD is a square. [A rhombus is a square if its diagonals are equal]

Question 16. Find the co-ordinates of circumcentre and radius of circumcircle of AABC if A (7, 1), B (3,5) and C (2,0) are given.

Solution & Step-by-Step Answer:
Suppose, O (h, k) is the circumcentre of ∆ABC ∴ h2 – 6h + 9 + k2 – 10k + 25 = h2 – 4h + 4 + k2 ∴ 2h + 10k = 30 ∴ h + 5k = 15 … (ii)[Dividing both sides by 2] Multiplying equation (i) by 5, we get 25h + 5k = 115 …(iii) Subtracting equation (ii) from (iii), we get Substituting the value of h in equation (i), we get ∴ The co-ordinates of the circumcentre of the triangle are (,) and radius of circumcircle is units.
Question 17 Maharashtra Board Solution
Given A (4, -3), B (8, 5). Find the co-ordinates of the point that divides segment AB in the ratio 3:1.
Solution & Step-by-Step Answer:
Suppose point C divides seg AB in the ratio 3:1. Here; A(x1, y1) = A (4, -3) B (x2, y2) = B (8, 5) By section formula, ∴ The co-ordinates of point dividing seg AB in ratio 3 : 1 are (7, 3).
Question 18 Maharashtra Board Solution
Find the type of the quadrilateral if points A (-4, -2), B (-3, -7), C (3, -2) and D (2, 3) are joined serially.
Solution & Step-by-Step Answer:
Slope of AB = slope of CD ∴ line AB || line CD slope of BC = slope of AD ∴ line BC || line AD Both the pairs of opposite sides of ∆ABCD are parallel. ∴ ꠸ ABCD is a parallelogram. ∴ The quadrilateral formed by joining the points A, B, C and D is a parallelogram.
Question 19 Maharashtra Board Solution
The line segment AB is divided into five congruent parts at P, Q, R and S such that A-P-Q-R-S-B. If point Q (12, 14) and S (4, 18) are given, find the co-ordinates of A, P, R, B.
Solution & Step-by-Step Answer:
Points P, Q, R and S divide seg AB in five congruent parts. Let A (x1, y1), B (x2, y2), P (x3, y3) and R (x4, y4) be the given points. Point R is the midpoint of seg QS. By midpoint formula, x co-ordinate of R = = = 8 y co-ordinate of R = = = 16 ∴ co-ordinates of R are (8, 16). Point Q is the midpoint of seg PR. By midpoint formula, ∴ 28 = y3 + 16 ∴ y3 = 12 ∴ P(x3,y3) = (16, 12) ∴ co-ordinates of P are (16, 12). Point P is the midpoint of seg AQ. By midpoint formula, ∴ co-ordinates of A are (20, 10). Point S is the midpoint of seg RB. By midpoint formula, ∴ 36 = y2 + 16 ∴ y2 = 20 ∴ B(x2, y2) = (0, 20) ∴ co-ordinates of B are (0, 20). ∴ The co-ordinates of points A, P, R and B are (20, 10), (16, 12), (8, 16) and (0, 20) respectively.
Question 20 Maharashtra Board Solution
Find the co-ordinates of the centre of the circle passing through the points P (6, -6), Q (3, -7) and R (3,3).
Solution & Step-by-Step Answer:
Suppose O (h, k) is the centre of the circle passing through the points P, Q and R. ∴ (h – 6)2 + (k + 6)2 = (h – 3)2 + (k + 7)2 ∴ h2 – 12h + 36 + k2 + 12k + 36 = h2 – 6h + 9 + k2 + 14k + 49 ∴ 6h + 2k = 14 ∴ 3h + k = 7 …(i)[Dividing both sides by 2] OP = OR …[Radii of the same circle] ∴ (h – 6)2 + (k + 6)2 = (h – 3)2 + (k – 3)2 ∴ h2 – 12h + 36 + k2 + 12k + 36 = h2 – 6h + 9 + k2 – 6k + 9 ∴ 6h – 18k = 54 ∴ 3h – 9k = 27 …(ii)[Dividing both sides by 2] Subtracting equation (ii) from (i), we get Substituting the value of k in equation (i), we get 3h – 2 = 7 ∴ 3h = 9 ∴ h = = 3 ∴ The co-ordinates of the centre of the circle are (3, -2).
Question 21 Maharashtra Board Solution
Find the possible pairs of co-ordinates of the fourth vertex D of the parallelogram, if three of its vertices are A (5, 6), B (1, -2) and C (3, -2).
Solution & Step-by-Step Answer:
Let the points A (5, 6), B (1, -2) and C (3, -2) be the three vertices of a parallelogram. The fourth vertex can be point D or point Di or point D2 as shown in the figure. Let D(x1,y1), D, (x2, y2) and D2 (x3,y3). Consider the parallelogram ACBD. The diagonals of a parallelogram bisect each other. ∴ midpoint of DC = midpoint of AB Co-ordinates of point D(x1, y1) are (3, 6). Consider the parallelogram ABD1C. The diagonals of a parallelogram bisect each other. ∴ midpoint of AD1 = midpoint of BC ∴ Co-ordinates of D1(x2,y2) are (-1,-10). Consider the parallelogram ABCD2. The diagonals of a parallelogram bisect each other. ∴ midpoint of BD2 = midpoint of AC ∴ co-ordinates of point D2 (x3, y3) are (7, 6). ∴ The possible pairs of co-ordinates of the fourth vertex D of the parallelogram are (3, 6), (-1,-10) and (7,6).
Question 22 Maharashtra Board Solution
Find the slope of the diagonals of a quadrilateral with vertices A (1, 7), B (6,3), C (0, -3) and D (-3,3).
Solution & Step-by-Step Answer:
Suppose ABCD is the given quadrilateral. ∴ The slopes of the diagonals of the quadrilateral are 10 and 0.
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