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## Class 10 Maths MCQs Chapter 7 Coordinate Geometry

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

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 = \(\sqrt{(4-1)^{2}+(0-4)^{2}}=\sqrt{9+16}=\sqrt{25}=5\)

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

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 = \(\frac{1}{2}\) [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

= \(\frac{1}{2}\)[-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

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) = \(\sqrt{(\alpha-0)^{2}+(\beta-0)^{2}}=\sqrt{\alpha^{2}+\beta^{2}}\)

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

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

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.

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

**Answer/ Explanation**

Answer: c

Explaination:

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:

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

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

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

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)

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.

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

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:

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:

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:

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:

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:

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

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:

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:

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:

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:

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:

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:|

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

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

31. The area (in square units) of the triangle formed by the points A(a, 0), 0(0, 0) and B(0, b) is

**Answer/Explanation**

Answer: b

Explaination:

A (a, 0), O(0, 0) and B(0, b)

ar(∆AOB) = \(\frac{1}{2}\)|a(0 – b) + 0(b – 0) + 0(0 -0)|

= \(\frac{1}{2}\)|-ab + 0 + 0| = \(\frac{ab}{2}\)

32. The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the figure is [NCERT Exemplar Problems]

**Answer/Explanation**

Answer: a

Explaination:

∵ AOB is a right triangle.

∴ Mid-point of AB is equidistant from A, O and B.

33. The vertices of a triangle are (0, 0), (3, 0) and (0, 4). The centroid of the triangle is

**Answer/Explanation**

Answer: b

Explaination:

34. If A = (a , 2a) and B = \(\left(\frac{1}{a^{2}},-\frac{2}{a}\right)\) S = (1,0), then \(\frac{1}{\mathrm{SA}}+\frac{1}{\mathrm{SB}}=\) = _______ .

**Answer/Explanation**

Answer:

Explaination:

35. Find the distance of a point P(x, y) from the origin. [CBSE 2018]

**Answer/Explanation**

Answer:

Explaination:

36. Find the distance between the points \(\left(-\frac{8}{5}, 2\right) \text { and }\left(\frac{2}{5}, 2\right)\)

**Answer/Explanation**

Answer:

Explaination:

37. If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k? [Delhi 2017]

**Answer/Explanation**

Answer:

Explaination:

38. Show that (1, -1) is the centre of the circle circumscribing the triangle whose angular points are (4, 3), (- 2, 3) and (6, – 1).

**Answer/Explanation**

Answer:

Explaination:

39. A circle has its centre at the origin and a point P(5, 0) lies on it. The point Q(6, 8) lies outside the circle. State whether true or false. Justify your answer. [NCERT Exemplar Problems]

**Answer/Explanation**

Answer:

Explaination:

True, because distance between centre (origin) and Q(6,8) is greater than its radius, i.e. 5.

40. Find the coordinates of a point A, where AB is diameter of a circle whose centre is (2, -3) and B is the point (1, 4).

[Delhi 2019]

**Answer/Explanation**

Answer:

Explaination:

AB is diameter of the circle.

Let C be centre of circle, coordinates of C are (2, -3). So, C is mid-point of AB (diameter).

Let coordinates of A are (x, y).

41. A straight line is drawn joining the points (3, 4) and (5, 6). If the line is extended, the ordinate of the point on the line, whose abscissa is -1 is ________ .

**Answer/Explanation**

Answer:

Explaination:

42. In figure, P(5, -3) and Q(3,y) are the points of trisection of the line segment joining A(7, -2) and B (1, -5). Then y equals ________ .

**Answer/Explanation**

Answer:

Explaination:

43. A(5, 1); B(l, 5) and C(-3, – 1) are the vertices of ∆ABC. Find the length of median AD. [CBSE 2018 (C)]

**Answer/Explanation**

Answer:

Explaination:

AD is median of ΔABC

∴ D is mid-point of BC

Coordinates of

44. If the mid-point of the line segment joining the points P(6, b – 2) and Q(- 2, 4) is (2, – 3), find the value of b.

**Answer/Explanation**

Answer:

Explaination:

45. If A(1, 2), B(4, 3) and C(6, 6) are the three vertices of a parallelogram ABCD, find the coordinates of the fourth vertex D.

**Answer/Explanation**

Answer:

Explaination:

46. Find the coordinates of the centroid of a triangle whose vertices are (0, 6), (8, 12) and (8, 0).

**Answer/Explanation**

Answer:

Explaination:

47. Two vertices of a triangle are (3, -5) and (-7, 4). If its centroid is (2, – 1), find the third vertex.

**Answer/Explanation**

Answer:

Explaination:

48. The coordinates of one end point of a diameter of a circle are (4, -1) and the coordinates of the centre are (1, -3). Find the coordinates of the other end of the diameter.

**Answer/Explanation**

Answer:

Explaination:

Given that coordinates of one end point of the diameter is (4, -1) and centre of the circle is (1, – 3).

Let coordinates of the other end of the diameter be (x, y).

We know that the centre of the circle (1, -3) is the mid-point of diameter.

⇒ \(\frac{4+x}{2}\) = 1 and \(\frac{(-1+y)}{2}\) ,

⇒ 4 + x = 2 and -1 + y = -6

⇒ x = – 2 and y = -6 + 1 = -5

Thus, coordinates of the other end of the diameter are (-2, -5).

49. Point P divides the line segment joining the points A(2, -5) and B(5, 2) in the ratio 2:3. Name the quadrant in which P lies. [Delhi 2011]

**Answer/Explanation**

Answer:

Explaination:

50. In figure, P(5, -3) and Q(3, vj are the points of trisection of the line segment joining A(7, -2) and B(1, -5). Find y [Foreign 2013]

**Answer/Explanation**

Answer:

Explaination:

51. Area of the triangle formed by (1, – 4), (3, – 2) and (- 3, 16) is _______ .

**Answer/Explanation**

Answer:

Explaination:

Area of the ∆ = \(\frac{1}{2}\)|1(-2 – 16) + 3(16 + 4) + (-3) (-4 + 2)|

= \(\frac{1}{2}\)| -18 + 60 + 6 | = 24 sq. units

52. If the points (- 2, – 5), (2, – 2) and (8, p) are collinear, then the value of p is _______ .

**Answer/Explanation**

Answer:

Explaination:

Points are collinear

∴ x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1} – x_{2}y_{1}– x_{3}y_{2} – x_{1}y_{3} = 0

⇒ (-2) × (-2) + (2) × (p) + 8 × (-5) – (2) × (-5) – (8) × (-2) – (-2) ×(p) = 0

⇒ 4 + 2p – 40 + 10 + 16 + 2p = 0

⇒ 4p – 40 + 30 = 0

⇒ 4p = 10 10 5

⇒ p = \(\frac{10}{4}\) = \(\frac{5}{2}\)/

53. Show that the points P(\(\frac{-3}{2}\), 3), Q(6, -2) and R(-3, 4) are collinear.

**Answer/Explanation**

Answer:

Explaination:

If [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] = 0 then points are collinear.

∴ \(\frac{-3}{2}\)(-2 – 4) + 6(4 – 3) + [(-3){3-(-2)}]

= \(\frac{-3}{2}\) × (-6) + 6 × 1 – 3 × 5

= 9 + 6 – 15 = 0.

∴ P, Q and R are collinear.

54. Find the area (in square units) of the triangle formed by the points A(a, 0), 0(0, 0) and B(0, b). [Foreign 2011]

**Answer/Explanation**

Answer:

Explaination:

A (a, 0), O(0, 0) and B(0, b)

ar(∆AOB) = \(\frac{1}{2}\)|a(0 – b) + 0{b – 0) + 0(0 -0)|

= \(\frac{1}{2}\)| -ab + 0 +0| = \(\frac{ab}{2}\)

55. In figure, find the area of triangle ABC.

**Answer/Explanation**

Answer:

Explaination:

56. If area of the triangle given below is 20 square units, what are the coordinates of point C

**Answer/Explanation**

Answer:

Explaination:

Drow AD ⊥ OC

∴ AD = b

and coordinates of C are (h, 0)

∴ OC = h

Area of ∆OAC = 20 sq. units

⇒ \(\frac{1}{2}\) × OC × AD = 20

h × b = 40

h = \(\frac{40}{b}\)

∴ Coordinates of C are (\(\frac{40}{b}\), 0)

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