Practice Test: Coordinate Geometry - Class 10 MCQ

# Practice Test: Coordinate Geometry - Class 10 MCQ

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## 25 Questions MCQ Test Mathematics (Maths) Class 10 - Practice Test: Coordinate Geometry

Practice Test: Coordinate Geometry for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The Practice Test: Coordinate Geometry questions and answers have been prepared according to the Class 10 exam syllabus.The Practice Test: Coordinate Geometry MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Coordinate Geometry below.
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Practice Test: Coordinate Geometry - Question 1

### The distance between the points (a, b) and (– a, – b) is :

Detailed Solution for Practice Test: Coordinate Geometry - Question 1

We have distance formula as d =
Where x1=a,y1=b,x2=-a,y2=-b

Practice Test: Coordinate Geometry - Question 2

### The distance between points (a + b, b + c) and (a – b, c – b) is :

Detailed Solution for Practice Test: Coordinate Geometry - Question 2
Distance between two point is
root[{(a+b)-(a-b)}^2] + [{(b+c)-(c-b)}^2]
=root [{(a+b-a+b)^2}] + [((b+c-c+b)}^2]
=root[(2b)^2 + (2b)^2]
=root[4b^2 +4b^2]
=root[8b^2]
=2.root(2)b
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Practice Test: Coordinate Geometry - Question 3

### The distance between A (1, 3) and B (x, 7) is 5. The value of x > 0 is :

Practice Test: Coordinate Geometry - Question 4

The distance between the points (a cos 20° + b sin 20°, 0) and (0, a sin 20° – b cos 20°) is :

Practice Test: Coordinate Geometry - Question 5

Mid-point of the line-segment joining the points (– 5, 4) and (9, – 8) is :

Practice Test: Coordinate Geometry - Question 6

The co-ordinates of the points which divides the join of (– 2, – 2) and (– 5, 7) in the ratio 2 : 1 is :

Practice Test: Coordinate Geometry - Question 7

The co-ordinates of the point on x-axis which is equidistant from the points (5, 4) and (– 2, 3) are :

Detailed Solution for Practice Test: Coordinate Geometry - Question 7

Let the point on x-axis be (a,0)
Since PA = PB  PA2 = PB2
(a + 2)2 + (0 - 5)2 = (a - 2)2 + (0 + 3)2
(a + 2)2 - (a - 2)2 = 9 - 25 = -16
8a = -16  a = -2.
The required point is (-2,0).

Practice Test: Coordinate Geometry - Question 8

The co-ordinates of the point on y-axis which is equidistant from the points (3, 1) and (1, 5) are:

Practice Test: Coordinate Geometry - Question 9

The coordinates of the centre of a circle are (– 6, 1.5). If the ends of a diameter are (– 3, y) and (x, – 2) then:

Practice Test: Coordinate Geometry - Question 10

The points (– 2, 2), (8, – 2) and (– 4, – 3) are the vertices of a :

Practice Test: Coordinate Geometry - Question 11

The points (1, 7), (4, 2), (– 1, 1) and (– 4, 4) are the vertices of a :

Detailed Solution for Practice Test: Coordinate Geometry - Question 11

Step-by-step explanation:

Given points are (1,7),(4,2),(-1,-1),(-4,4)

Let the points are A,B,C,D.

Distance formula = √(x₂-x₁)²+(y₂-y₁)²

AB = √(4-1)²+(2-7)²

= √9+25

= √34

BC = √(-1-4)²+(-1-2)²

= √25+9

=  √34

CD = √(-4-(-1))²+(4-(-1))²

= √9+25

=  √34

DA = √(1-(-4))²+(7-4)²

= √25+9

= √34

also diagonals AC and BD

AC = √(1+1)² + (7+1)² = √68

BD = √(4+4)² + (2-4)² = √68

We know that the all sides and both diagonals of the square are equal

So,from this we can say that these points are the vertices of a square.

Practice Test: Coordinate Geometry - Question 12

The line segment joining (2, – 3) and (5, 6) is divided by x-axis in the ratio:

Detailed Solution for Practice Test: Coordinate Geometry - Question 12

Hence, the ratio is 1:2 and the division is internal.

Practice Test: Coordinate Geometry - Question 13

The line segment joining the points (3, 5) and (– 4, 2) is divided by y-axis in the ratio:

Detailed Solution for Practice Test: Coordinate Geometry - Question 13

Practice Test: Coordinate Geometry - Question 14

If (3, 2), (4, k) and (5, 3) are collinear then k is equal to :

Practice Test: Coordinate Geometry - Question 15

If the points (p, 0), (0, q) and (1, 1) are collinear then 1/p+1/q is equal to :

Practice Test: Coordinate Geometry - Question 16

Two vertices of a triangle are (–2, – 3) and (4, –1) and centroid is at the origin. The coordinates of the third vertex of the triangle are :

Practice Test: Coordinate Geometry - Question 17

A (5, 1), B(1, 5) and C (–3, –1) are the vertices of ?ABC. The length of its median AD is :

Practice Test: Coordinate Geometry - Question 18

Three consecutive vertices of a parallelogram are (1, –2), (3, 6) and (5, 10). The coordinates of the fourth vertex are :

Practice Test: Coordinate Geometry - Question 19

The vertices of a parallelogram are (3, –2), (4, 0), (6, –3) and (5, –5). The diagonals intersect at the point M. The coordinates of the point M are :

Detailed Solution for Practice Test: Coordinate Geometry - Question 19

Practice Test: Coordinate Geometry - Question 20

If two vertices of a parallelogram are (3, 2) and (–1, 0) and the diagonals intersect at (2, –5), then the other two vertices are :

Practice Test: Coordinate Geometry - Question 21

The circumcentre of the triangle formed by the lines xy + 2x + 2y + 4 = 0 and x + y + 2 = 0 is :

Practice Test: Coordinate Geometry - Question 22

The vertices of a triangle are (a, b – c), (b, c – a) and (c, a – b), then it's centroid lies on :

Practice Test: Coordinate Geometry - Question 23

The points (1, 2), (3, 8) and (x, 20) are collinear if x =

Detailed Solution for Practice Test: Coordinate Geometry - Question 23

Practice Test: Coordinate Geometry - Question 24

For the triangle whose sides are along the lines x = 0, y = 0 and x/6+y/8 = 1, the incentre is :

Practice Test: Coordinate Geometry - Question 25

For the triangle whose sides are along the lines y = 15, 3x – 4y = 0, 5x + 12y = 0, the incentre is :

Detailed Solution for Practice Test: Coordinate Geometry - Question 25

Given equations:

3x – 4y = 0 …(1)

5x+12y = 0 …(2)

Y-15 = 0 …(3)

From the given equations, (1), (2) and (3) represent the sides AB, BC and CA respectively.

Solving (1) and (2), we get

x= 0, and y= 0

Therefore, the side AB and BC intersect at the point B (0, 0)

Solving (1) and (3), we get

x= 20, y= 15

Hence, the side AB and CA intersect at the point A (20, 15)

Solving (2) and (3), we get

x= -36, y = 15

Thus, the side BC and CA intersect at the point C (-36, 15)

Now,

BC = a = 39
CA = b = 56
AB = c = 25
Similarly, (x1, y1) = A(20, 15)

(x2, y2) = B(0, 0)
(x3, y3) = C(-36, 15)

Therefore, incentre is

## Mathematics (Maths) Class 10

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## Mathematics (Maths) Class 10

116 videos|420 docs|77 tests