Class 10-Coordinate Geometry MCQs Multiple Choice Questions with Answers

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Class 10-Coordinate Geometry MCQ Questions with Answers

1. The distance of the point P(2, 3) from the x-axis is
(a) 2
(b) 3
(c) 1
(d) 5

Answer/ Explanation

Answer: b
Explaination: Reason: The distance from x-axis is equal to its ordinate i.e., 3

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2. The distance between the point P(1, 4) and Q(4, 0) is
(a) 4
(b) 5
(c) 6
(d) 3√3

Answer/ Explanation

Answer: b
Explaination: Reason: The required distance = \(\sqrt{(4-1)^{2}+(0-4)^{2}}=\sqrt{9+16}=\sqrt{25}=5\)

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3. The points (-5, 1), (1, p) and (4, -2) are collinear if
the value of p is
(a) 3
(b) 2
(c) 1
(d) -1

Answer/ Explanation

Answer: d
Explaination: Reason: The points are collinear if area of Δ = 0
= \(\frac{1}{2}\)[-5(p + 2) +l(-2 -1) + 4(1 – p)] – 0
⇒ -5 p -10-3 + 4-4p = 0
⇒ -9p = +9
∴ p = -1

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4. The area of the triangle ABC with the vertices A(-5, 7), B(-4, -5) and C(4, 5) is
(a) 63
(b) 35
(c) 53
(d) 36

Answer/ Explanation

Answer: c
Explaination: Reason: Area of ΔABC = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\)[-5(-5 – 5) -4(5 – 7) + 4(7 – (-5))] = \(\frac{1}{2}\)[-5(-10) -4(-2) + 4(12)]
= \(\frac{1}{2}\)[50 + 8 + 48] = \(\frac{1}{2}\) × 106 = 53 sq. units

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5. The distance of the point (α, β) from the origin is
(a) α + β
(b) α² + β²
(c) |α| + |β|
(d) \(\sqrt{\alpha^{2}+\beta^{2}}\)

Answer/ Explanation

Answer: d
Explaination: Reason: Distance of (α, β) from origin (0, 0) = \(\sqrt{(\alpha-0)^{2}+(\beta-0)^{2}}=\sqrt{\alpha^{2}+\beta^{2}}\)

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6. The area of the triangle whose vertices are A(1, 2), B(-2, 3) and C(-3, -4) is
(a) 11
(b) 22
(c) 33
(d) 21

Answer/ Explanation

Answer: a
Explaination: Reason: Required area= \(\frac{1}{2}\)[1(3 + 4) -2(-4 – 2) -3(2 – 3)]
= \(\frac{1}{2}\)[7 + 12 + 3]
= \(\frac{1}{2}\) × 22 = 11

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7. The line segment joining the points (3, -1) and (-6, 5) is trisected. The coordinates of point of trisection are
(a) (3, 3)
(b) (- 3, 3)
(c) (3, – 3)
(d) (-3,-3)

Answer/ Explanation

Answer: b
Explaination: Reason: Since the line segment AB is trisected


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8. The line 3x + y – 9 = 0 divides the line joining the points (1, 3) and (2, 7) internally in the ratio
(a) 3 : 4
(b) 3 : 2
(c) 2 : 3
(d) 4 : 3

Answer/ Explanation

Answer: a
Explaination: Reason: Let the line 3x + y – 9 = 0 divide the line segment joining A(l, 3) ad B(2, 7) in the ratio K : 1 at point C.


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9. The distance between A (a + b, a – b) and B(a – b, -a – b) is

Answer/ Explanation

Answer: c
Explaination:


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10. If (a/3, 4) is the mid-point of the segment joining the points P(-6, 5) and R(-2, 3), then the value of ‘a’ is
(a) 12
(b) -6
(c) -12
(d) -4

Answer/ Explanation

Answer: c
Explaination:


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11. If the distance between the points (x, -1) and (3, 2) is 5, then the value of x is
(a) -7 or -1
(b) -7 or 1
(c) 7 or 1
(d) 7 or -1

Answer/ Explanation

Answer: d
Explaination: Reason: We have \(\sqrt{(x-3)^{2}+(-1-2)^{2}}=5\)
⇒ (x – 3)² + 9 = 25
⇒ x² – 6x + 9 + 9 = 25
⇒ x² -6x – 7 = 0
⇒ (x – 7)(x + 1) = 0
⇒ x = 7 or x = -1

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12. The points (1,1), (-2, 7) and (3, -3) are
(a) vertices of an equilateral triangle
(b) collinear
(c) vertices of an isosceles triangle
(d) none of these

Answer/ Explanation

Answer: b
Explaination: Reason: Let A(1, 1), B(-2, 7) and C(3, 3) are the given points, Then, we have


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13. The coordinates of the centroid of a triangle whose vertices are (0, 6), (8,12) and (8, 0) is
(a) (4, 6)
(b) (16, 6)
(c) (8, 6)
(d) (16/3, 6)

Answer/ Explanation

Answer: d
Explaination: Reason: The co-ordinates of the centroid of the triangle is


 

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14. Two vertices of a triangle are (3, – 5) and (- 7,4). If its centroid is (2, -1), then the third vertex is
(a) (10, 2)
(b) (-10,2)
(c) (10,-2)
(d) (-10,-2)

Answer/ Explanation

Answer: c
Explaination: Reason: Let the coordinates of the third vertex be (x, y)


 

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15. The area of the triangle formed by the points A(-1.5, 3), B(6, -2) and C(-3, 4) is
(a) 0
(b) 1
(c) 2
(d) 3/2

Answer/ Explanation

Answer: a
Explaination: Reason: Area of ΔABC = \(\frac{1}{2}\) [-1.5(-2 – 4) + 6(4 – 3) + (-3) (3 + 2)] = \(\frac{1}{2}\) [9 + 6 – 15] = 0. It is a straight line.

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16. If the points P(1, 2), B(0, 0) and C(a, b) are collinear, then
(a) 2a = b
(b) a = -b
(c) a = 2b
(d) a = b

Answer/ Explanation

Answer: a
Explaination: Reason: Area of ΔPBC = 0
⇒ \(\frac{1}{2}\)[1(0 – b) + 0(6 – 1) + a(2 – 0)] = 0
⇒ \(\frac{1}{2}\)[-6 + 2a] = 0
⇒ -b + 2a = 0
∴ 2a = b

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17. A triangle with vertices (4, 0), (- 1, – 1) and (3, 5) is a/an
(a) equilateral triangle
(b) right-angled triangle
(c) isosceles right-angled triangle
(d) none of these

Answer/Explanation

Answer: a
Explaination:


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18. The points (- 4, 0), (4, 0) and (0, 3) are the vertices of a/an [NCERT Exemplar Problems]
(a) right triangle
(b) isosceles triangle
(c) equilateral triangle
(d) scalene triangle

Answer/Explanation

Answer: b
Explaination:


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19. 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 [NCERT Exemplar Problems]


Answer/Explanation

Answer: d
Explaination:


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20. If the distance between the points(4, p) and (1, 0) is 5 units, then the value of p is [NCERT Exemplar Problems]
(a) 4 only
(b) ± 4
(c) -4 only
(d) 0

Answer/Explanation

Answer: b
Explaination:


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21. The points (2, 5), (4, – 1) and (6, – 7) are vertices of an/a
(a) isosceles triangle
(b) equilateral triangle
(c) right-angled triangle
(d) none of these

Answer/Explanation

Answer: d
Explaination:



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22. If the segment joining the points (a, b) and (c, d) subtends a right angle at the origin, then
(a) ac – bd = 0
(b) ac + bd = 0
(c) ab + cd = 0
(d) ab – cd= 0

Answer/Explanation

Answer: b
Explaination:
Let A {a, b), B(c, d), 0(0, 0)
∴ ∠AOB = 90°
⇒ AB² = AO² + BO²
(c – a)² + (d- b)² = a² + b² + c² + d²
⇒ ac + bd = 0

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23. AOBC is a rectangle whose three vertices are A(0, 3), 0(0, 0) and B(5, 0). The length of its diagonal is [NCERT Exemplar Problems]
(a) 5
(b) 3
(c) √34
(d) 4

Answer/Explanation

Answer: c
Explaination:


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24. The perimeter of a triangle with vertices (0,4), (0, 0) and (3, 0) is [NCERT Exemplar Problems]
(a) 5
(b) 12
(c) 11
(d) 7 + √5

Answer/Explanation

Answer: b
Explaination:
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25. If the distance between the points (4, p) and (1, 0) is 5 units, then the value of p is [NCERT Exemplar Problems]
(a) 4 only
(b) ±4
(c) -4 only
(d) 0

Answer/Explanation

Answer: b
Explaination:


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26. If p\(\left(\frac{a}{3}, 4\right)\) is the mid-point of the line segment joining the points Q (-6, 5) and R (-2, 3), then the value of a is [NCERT Exemplar Problems]
(a) -4
(b) -12
(c) 12
(d) -6

Answer/Explanation

Answer: b
Explaination:


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27. If P(l, 2), Q(4, 6), R(5, 7) and S(a, b) are the vertices of a parallelogram PQRS, then
(a) a = 2,b = A
(b) a = 3,b = 4
(c) a = 2, b = 3
(d) a = 3, b = 5

Answer/Explanation

Answer: c
Explaination:


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28. 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/Explanation

Answer: d
Explaination:|


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29. The points (k + 1, 1), (2k + 1, 3) and (2k + 2, 2k) are collinear if
(a) k = -1, 2
(b) k=\(\frac{1}{2}\),2
(c) k = 2, 1
(d) k = –\(\frac{1}{2}\),2

Answer/Explanation

Answer: d
Explaination:
∵ Points are collinear.
∴ (k + 1) (3 – 2k) + (2k + 1) (2k- 1) + (2k + 2) (1 – 3) = 0
⇒ 3k + 3 – 2k² – 2k + 4k² – 1 -4k – 4 = 0
⇒ 2k² – 3k – 2 = 0
⇒ 2k² – 4k + k – 2 = 0
⇒ 2k(k – 2) + 1(k – 2) = 0
⇒ (2k + 1) (k – 2) = 0
k = –\(\frac{1}{2}\), k = 2

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30. The area of the triangle with vertices at the points (a, b + c), (b, c + a) and (c, a + b) is
(a) (a + b + c) sq. units
(b) (a + b – c) sq. units
(c) (a – b + cj sq. units
(d) 0

Answer/Explanation

Answer: d
Explaination:
Using formula for area of triangle, we get
Area = zero.
Area of triangle
= \(\frac{1}{2}\)| a(c + a- a-b) + b(a+ b- b-c) + c(b + c- c-a) |
= \(\frac{1}{2}\) |ac – ab + ba – bc + cb – ca| = 0

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