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Explore various MCQ Questions for Class 10 Maths Coordinate Geometry with answers provided with detailed solutions by looking below: -
Practice Class 10 Maths MCQ Question of Coordinate geometry
1. The horizontal and vertical lines drawn to determine the position of a point in a Cartesian plane are called
(a) Intersecting lines
(b) Transversals
(c) Perpendicular lines
(d) X-axis and Y-axis
2. 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
3. The distance of the point P (2, 3) from the x-axis is
(a) 2
(b) 3
(c) 1
(d) 5
4. The points (3, 2), (0, 5), (-3, 2) and (0, -1) are the vertices of a quadrilateral. Which quadrilateral is it?
(a) Rectangle
(b) Square
(c) Parallelogram
(d) Rhombus
5. The distance between the point P(1, 4) and Q(4, 0) is
(a) 4
(b) 5
(c) 6
(d) 3√3
6. 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
7. 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
8. 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
9. 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
10. 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
11. The distance of a point from the y axis is called its ———
(a) Ordinate
(b) Abscissa
(c) Origin
(d) None of these
12. The distance of a point from the x axis is called its —————
(a) Ordinate
(b) Abscissa
(c) Origin
(d) None of these
13. The midpoints of a line segment joining two points A(2, 4) and B(-2, -4)
(a) (-2,4)
(b) (2,-4)
(c) (0, 0)
(d) (-2,-4)
14. If O(p/3, 4) is the midpoint of the line segment joining the points P(-6, 5) and Q(-2, 3). The value of p is:
(a) 7/2
(b) -12
(c) 4
(d) -4
15. The points which divides the line segment of points P(-1, 7) and (4, -3) in the ratio of 2:3 is:
(a) (-1, 3)
(b) (-1, -3)
(c) (1, -3)
(d) (1, 3)
16. The ratio in which the line segment joining the points P(-3, 10) and Q(6, – 8) is divided by O(-1, 6) is:
(a) 1:3
(b) 3:4
(c) 2:7
(d) 2:5
17. The coordinates of a point P, where PQ is the diameter of circle whose centre is (2, – 3) and Q is (1, 4) is:
(a) (3, -10)
(b) (2, -10)
(c) (-3, 10)
(d) (-2, 10)
18. A linear equation in two variables of the form ax + by + c = 0, when represented graphically gives a ——————
(a) Parabola
(b) Circle
(c) Straight line
(d) None of these
19. The graph of a quadratic equation is a ———
(a) Parabola
(b) Circle
(c) Straight line
(d) None of these
20. 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
Answers & Explanations
1. Answer: (d) X-axis and Y-axis
2. Answer: (b) collinear
3. Answer: (b) 3
Explanation: The distance from x-axis is equal to its ordinate i.e., 3
4. Answer: (b) Square
5. Answer: (b) 5
Explanation: The required distance = \(\sqrt{(4-1)^2+(0-4)^2}\)
=\(\sqrt{9+16}\)
=\(\sqrt{25}\)
=5
6. Answer: (d) -1
Explanation: 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
7. Answer: (a) 11
Explanation: 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
8. Answer: (d) 7 or -1
Explanation: We have \(\sqrt{(x−3)^2+(−1−2)^2} =5\)
⇒ (x – 3)\(^2\) + 9 = 25
⇒ x\(^2\) – 6x + 9 + 9 = 25
⇒ x\(^2\) -6x – 7 = 0
⇒ (x – 7)(x + 1) = 0
⇒ x = 7 or x = -1
9. Answer: (a) 0
Explanation: 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.
10. Answer: (a) 2a = b
Explanation: 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
11. Answer: (b) Abscissa
12. Answer: (a) Ordinate
13. Answer: (c) (0, 0)
Explanation: As per midpoint formula, we know;
x=[2+(-2)]/2 = 0/2 = 0
y=[4+(-4)]/2=0/2=0
Hence, (0,0) is the midpoint of of AB.
14. Answer: (b)-12
Explanation: Since, (p/3, 4) is the midpoint of line segment PQ, thus;
p/3 = (-6-2)/2
p/3 = -8/2
p/3 = -4
p= -12
Therefore, the value of p is -12.
15. Answer: (d)(1, 3)
Explanation: By section formula we know:
x=[(2.4)+(3.(-1))]/(2+3) = (8-3)/5 = 1
y=[(2.(-3))+(3.7)]/(2+3) = (-6+21)/5 = 3
Hence, the required point is (1,3)
16. Answer: (c) 2:7
Explanation: Let the ratio in which the line segment joining P( -3, 10) and Q(6, -8) is divided by point O( -1, 6) be k :1.
So, -1 = ( 6k-3)/(k+1)
–k – 1 = 6k -3
7k = 2
k = 2/7
Hence, the required ratio is 2:7.
17. Answer: (a)(3, -10)
Explanation: By midpoint formula, we know;
[(x+1)/2,(y+4)/2] = (2,-3) (Since, O is the midpoint of PQ)
(x+1)/2 = 2
x+1=4
x=3
(y+4)/2 = -3
y+4=-6
y=-10
So, the coordinates of point P is (3, -10).
18. Answer: (c) Straight line
19. Answer: (a) Parabola
20. Answer: (d) 0
Explanation: 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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