CBSE Class 10 Mathematics Chapter 7 Coordinate Geometry Multiple Choice Questions with Answers. MCQ Class 10 Maths Coordinate Geometry with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Mathematics Coordinate Geometry MCQs with Answers to know their preparation level.
Multiple Choice Questions - Set - 5
Question 1:
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
Correct Answer – (C)
Question 2 :
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)
Correct Answer – (A)
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).
Question 3 :
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
Correct Answer – (B)
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.
Question 4 :
The area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order, is:
(a)12 sq.unit
(b)24 sq.unit
(c)30 sq.unit
(d)32 sq.unit
Correct Answer – (B)
To find the area of the rhombus, we need to find the length of its diagonals and use the below formula:
Area = ½ (Diagonal1)(Diagonal2)
Area = (1/2 ) (AC)(BD)
Diagonal1=√[(3-(-1))2+(0-4)2]= 4√2
Diagonal2 = √[(4-(-2))2+(5-(-1))2]=6√2
Area = ½ x 4√2 x 6√2 = 24 sq.unit.
Question 5 :
The distance of point A(2, 4) from x-axis is
(a)2
(b)4
(c)-2
(d)-4
Correct Answer – (B)
Question 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
Correct Answer – (D)
= 12[-5(p + 2) +l(-2 -1) + 4(1 – p)] – 0
⇒ -5 p -10-3 + 4-4p = 0
⇒ -9p = +9
∴ p = -1
Question 7 :
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)
Correct Answer – (D)
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)
Question 8 :
The distance between the points P(0, 2) and Q(6, 0) is
(a)4√10
(b)2√10
(c)√10
(d)20
Correct Answer – (B)
By distance formula we know:
PQ = √[(6-0)2+(0-2)2]
PQ = √(62+22)
PQ=√(36+4)
PQ=√40=2√10
Question 9 :
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
Correct Answer – (C)
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.
Question 10 :
The distance of the point P(2, 3) from the x-axis is
(a) 2
(b) 3
(c) 1
(d) 5
