Chapter 7: Coordinate Geometry

VVI questions — the ones the board asked most (20)

1. Q1Short answer2 marksAsked: 2022, 2026

   Prove that the points (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.

   

   Medha Ma’am

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   Model answer · distance formula, isosceles triangle

   1. 1.Let A(5, –2), B(6, 4) and C(7, –2) be the given points. The distance between P(x1, y1) and Q(x2, y2) is given by the distance formula, PQ = √[(x2 – x1)² + (y2 – y1)²]. Using this, AB = √[(6 – 5)² + (4 – (–2))²] = √[1 + 36] = √37. BC = √[(7 – 6)² + (–2 – 4)²] = √[1 + 36] = √37. AC = √[(7 – 5)² + (–2 – (–2))²] = √[4 + 0] = 2.1 mark
   2. 2.Since AB = BC = √37, two sides of the triangle ABC are equal in length. Therefore, A(5, –2), B(6, 4) and C(7, –2) are the vertices of an isosceles triangle.1 mark

   • Write the distance formula first before substituting the coordinates.
   • Clearly show AB = BC = √37 as the equal sides, and AC = 2 as the third side.
   • End with the statement that since AB = BC, the triangle is isosceles.

   Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme

2. Q2Short answer2 marksAsked: 2020, 2022

   Find the relation between x and y if the point P(x, y) is equidistant from the points A(1, 4) and B(-1, 2).

   

   Medha Ma’am

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   Hear it from Medha Ma’am · 85 s

   Model answer · relation between x and y, equidistant points

   1. 1.Let P(x, y) be equidistant from the points A(1, 4) and B(-1, 2). We are given that PA = PB. So, PA² = PB². Using the distance formula, PA² = (x – 1)² + (y – 4)² and PB² = (x + 1)² + (y – 2)².1 mark
   2. 2.Equating PA² and PB² obtained using the distance formula: (x – 1)² + (y – 4)² = (x + 1)² + (y – 2)² x² – 2x + 1 + y² – 8y + 16 = x² + 2x + 1 + y² – 4y + 4 – 2x + 1 – 8y + 16 = 2x + 1 – 4y + 4 – 4x – 4y + 12 = 0 x + y = 3 which is the required relation.1 mark

   • Begin by writing PA = PB and then PA² = PB² before expanding.
   • Show the full expansion of both squared terms using the distance formula as working.
   • End with the final simplified relation x + y = 3 clearly stated.

   Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme

3. Q3Short answer2 marksAsked: 2017, 2022

   Find the area of the triangle whose vertices are (1, -1), (-4, 6) and (-3, -5).

   

   Medha Ma’am

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   Hear it from Medha Ma’am · 90 s

   Model answer · area of triangle, coordinates

   1. 1.The area of the triangle formed by the vertices A(1, –1), B(–4, 6) and C(–3, –5) is given by the formula: Area = 1/2 |x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|.1 mark
   2. 2.Substituting the values x1 = 1, y1 = –1, x2 = –4, y2 = 6, x3 = –3, y3 = –5 in the formula, Area = 1/2 |1(6 − (−5)) + (−4)((−5) − (−1)) + (−3)((−1) − 6)| = 1/2 |1(11) + (−4)(−4) + (−3)(−7)| = 1/2 (11 + 16 + 21) = 1/2 (48) = 24. So, the area of the triangle is 24 square units.1 mark

   Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme

4. Q4Short answer2 marksAsked: 2017, 2022

   If the points A(2, 3), B(4, k) and C(6, -3) are collinear, then find the value of k.

   How the model answer begins

   1. If the points A(2, 3), B(4, k) and C(6, -3) are collinear, then the area of the triangle formed by these points must be zero. This is the collinearity condition for three points.

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5. Q5Short answer2 marksAsked: 2020, 2023

   Find the coordinates of a point on the y-axis which is equidistant from the points ( - 5, - 2 ) and ( 3, 2 ).

   How the model answer begins

   1. We know that a point on the y-axis is of the form (0, y). So, let the point P(0, y) be equidistant from the points A(–5, –2) and B(3, 2).

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6. Q6Long answer5 marksAsked: 2026

   If the points (-4, 8), (-3, -4), (0, -5) and (5, 6) are the vertices of a quadrilateral then find the area of the quadrilateral.

   How the model answer begins

   1. Let the vertices of the quadrilateral be A(–4, 8), B(–3, –4), C(0, –5) and D(5, 6), taken in order. Join B to D so that the quadrilateral ABCD is divided into two triangles ABD and BCD, which have no common area, so…

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7. Q7Long answer5 marksAsked: 2026

   Find the area of the quadrilateral whose vertices are ( 1, 1 ), ( 7, - 3 ), ( 12, 2 ) and ( 2, 3 ).

   How the model answer begins

   1. Join diagonal AC to divide quadrilateral ABCD [A(1, 1), B(7, −3), C(12, 2), D(2, 3)] into triangles ABC and ACD.

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8. Q8Long answer5 marksAsked: 2021

   Find the area of the triangle formed by joining the midpoints of the sides of the triangle with vertices (0, -1), (2, 1) and (0, 3).

   How the model answer begins

   1. The mid-point of the line segment joining the points P(x1, y1) and Q(x2, y2) is (x1+x2/2, y1+y2/2). The vertices of the given triangle are A(0, –1), B(2, 1) and C(0, 3). The mid-point of AB is (0+2/2, –1+1/2) = (1, 0).…

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9. Q9Short answer2 marksAsked: 2025

   Find the co-ordinates of the point which divides line segment joining the points (- 1, 7) and (4, - 3) in the ratio 2 : 3 internally.

   How the model answer begins

   1. The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2) internally in the ratio m1 : m2 are ((m1x2 + m2x1)/(m1+m2), (m1y2 + m2y1)/(m1+m2)). This is known as the…

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10. Q10Short answer2 marksAsked: 2025

    Find the area of the triangle whose vertices are (- 5, - 1), (3, - 5) and (5, 2).

    How the model answer begins

    1. The area of the triangle formed by the points (x1, y1), (x2, y2) and (x3, y3) is the numerical value of the expression 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)].

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11. Q11Short answer2 marksAsked: 2025

    For what value of k points (1, 1), (3, k) and (-1, 4) are collinear?

    How the model answer begins

    1. If the area of a triangle is 0 square units, then its vertices will be collinear. So for the points (1, 1), (3, k) and (-1, 4) to be collinear, the area of the triangle formed by them must be zero.

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12. Q12Short answer2 marksAsked: 2025

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

    How the model answer begins

    1. We know that a point on the y-axis is of the form (0, y). So, let the point P(0, y) be equidistant from A(6, 5) and B(– 4, 3), i.e., PA = PB.

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13. Q13Short answer2 marksAsked: 2026

    Find the distance between the points (4, 7) and (7, 3).

    How the model answer begins

    1. The distance between two points P(x1, y1) and Q(x2, y2) is given by the distance formula, PQ = √((x2 - x1)² + (y2 - y1)²).

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14. Q14Short answer2 marksAsked: 2026

    Prove that the points (1, 1), (-2, 7) and (3, -3) are collinear.

    How the model answer begins

    1. Points are collinear if the area of the triangle formed by them is zero. Let A(1, 1), B(-2, 7) and C(3, -3) be the given points. If the area of triangle ABC is zero, then A, B and C are collinear.

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15. Q15Short answer2 marksAsked: 2026

    Find the area of the triangle formed having vertices (-5, 6), (3, 0) and (9, 8).

    How the model answer begins

    1. The area of the triangle formed by the points (x1, y1), (x2, y2) and (x3, y3) is the numerical value of the expression 1/2 [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]. Here the vertices are A(-5, 6), B(3, 0) and C(9, 8),…

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16. Q16Short answer2 marksAsked: 2026

    Find the coordinates of a point P which divides the line segment joining the points A (-6, 10) and B (3, -8) internally in the ratio 2 : 7.

    How the model answer begins

    1. The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2) internally in the ratio m1 : m2 are given by the section formula: x = (m1x2 + m2x1) / (m1 + m2) and y =…

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17. Q17Short answer2 marksAsked: 2026

    Find a point on the x-axis which is equidistant from the points (-2, 5) and (2, -3).

    How the model answer begins

    1. The coordinates of a point on the x-axis are of the form (x, 0). So, let the required point on the x-axis be P(x, 0), which is equidistant from the points A(-2, 5) and B(2, -3).

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18. Q18Short answer2 marksAsked: 2026

    Find the area of the triangle, the coordinates of whose vertices are (5, 2), (4, 7) and (7, -4).

    How the model answer begins

    1. The area of the triangle formed by the vertices A(5, 2), B(4, 7) and C(7, –4) is given by the formula: area of the triangle is the numerical value of the expression 1/2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)].

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19. Q19Short answer2 marksAsked: 2026

    Prove that the points (7, -2), (5, 1) and (3, 4) are collinear.

    How the model answer begins

    1. Let the given points be A(7, -2), B(5, 1) and C(3, 4). The area of the triangle formed by these points is given by the formula: Area = 1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2)], where (x1, y1) = (7, -2), (x2, y2) = (5, 1)…

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20. Q20Short answer2 marksAsked: 2022

    Two vertices of a triangle ABC are A(6, 4) and B(-2, 2) and its centroid G is (3, 4). Find the coordinates of the third vertex C of triangle ABC.

    How the model answer begins

    1. If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of a triangle, the coordinates of the centroid of the triangle are given by the centroid formula: ((x1+x2+x3)/3, (y1+y2+y3)/3). Here A(6, 4), B(-2, 2) and C(x, y)…

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20 more questions of this chapter, each with its model answer, in the AskMedha app.

Objective questions with the board’s key (10)

1. 1. Bihar Board 2026

   Distance between the points (9, 3) and (15, 11) is

   • (A) 8 units
   • (B) 10 units ✓
   • (C) 12 units
   • (D) 15 units

   Answer: (B)

2. 2. Bihar Board 2026

   Perimeter of the triangle having vertices (0, 7), (0, 0) and (24, 0) is

   • (A) 28 units
   • (B) 42 units
   • (C) 48 units
   • (D) 56 units ✓

   Answer: (D)

3. 3. Bihar Board 2026

   Distance of the point (11, 60) from the origin is

   • (A) 51 units
   • (B) 53 units
   • (C) 59 units
   • (D) 61 units ✓

   Answer: (D)

4. 4. Bihar Board 2026

   If the points (1, 2), (0, 0) and (a, b) are collinear then

   • (A) a = b
   • (B) a = 2b
   • (C) 2a = b ✓
   • (D) a = -b

   Answer: (C)

5. 5. Bihar Board 2026

   Area of the triangle formed having vertices (3, 0), (7, 0) and (8, 4) is

   • (A) 16 cm^2
   • (B) 24 cm^2
   • (C) 8 cm^2 ✓
   • (D) 32 cm^2

   Answer: (C)

6. 6. Bihar Board 2026

   The length of the diagonal of a square is 10sqrt(2) cm; then its area is

   • (A) 200 cm^2
   • (B) 100 cm^2 ✓
   • (C) 150 cm^2
   • (D) 100sqrt(2) cm^2

   Answer: (B)

7. 7. Bihar Board 2026

   The distance between the points ( 8sin60°, 0 ) and ( 0, 8sin30°) ( in units ) is

   • (A) 6
   • (B) 7
   • (C) 8 ✓
   • (D) 4

   Answer: (C)

8. 8. Bihar Board 2026

   The point (- 6, 4) is located in which quadrant?

   • (A) First
   • (B) Second ✓
   • (C) Third
   • (D) Fourth

   Answer: (B)

9. 9. Bihar Board 2026

   Distance of the point P (- 6, - 15) from x-axis ( in units ) is

   • (A) 6
   • (B) 12
   • (C) 15 ✓
   • (D) 30

   Answer: (C)

10. 10. Bihar Board 2026

    Distance between the points (- 1, 3) and (- 5, 7) ( in units ) is

    • (A) 4 sqrt(2) ✓
    • (B) 3 sqrt(2)
    • (C) 2 sqrt(2)
    • (D) sqrt(2)

    Answer: (A)

Questions parents and students ask

How many marks does "Coordinate Geometry" carry in Bihar Board Class 10 Maths?

In the Bihar Board papers of 2021–2026, this chapter carried about 10 marks and about 14 questions a year; by marks it ranks 2 in Maths.

Which are the most important (VVI) questions of "Coordinate Geometry"?

The 20 questions above are from the board’s real papers, each with the years it was asked; the most-asked come first with their full model answers. All 40 written questions of this chapter, with answers, are in the AskMedha app.

Do the answers follow the board’s marking scheme?

BSEB does not publish a marking scheme. Each answer is written point by point on a structure derived from past papers and checked by a second, independent model; it is never called the board’s official scheme.