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Test: Distance Formula - Class 10 MCQ


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20 Questions MCQ Test Mathematics (Maths) Class 10 - Test: Distance Formula

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

The coordinates of the centre of a circle passing through (1, 2), (3, – 4) and (5, – 6) is:​

Detailed Solution for Test: Distance Formula - Question 1

 

Test: Distance Formula - Question 2

One end of a line of length 10 units is at the point (-3, 2). If the ordinate of the other end be 10, then the abscissa will be

Detailed Solution for Test: Distance Formula - Question 2

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Test: Distance Formula - Question 3

The horizontal and vertical lines drawn to determine the position of a point in a Cartesian plane are called​

Detailed Solution for Test: Distance Formula - Question 3

The point is determined on 2 dimensions .The name of horizontal and vertical lines drawn to determine the position of any point in the Cartesian plane is x-axis and y-axis respectively. The name of each part of the plane formed by these two lines x-axis and the y-axis is quadrants. The point where these two lines intersect is called the origin.

Test: Distance Formula - Question 4

Let P(x, y) be equidistant from the points A (7, 1) and (3, 5).Find a relation between x and y.​

Test: Distance Formula - Question 5

If the four points (0,-1), (6,7),(-2,3) and (8,3) are the vertices of a rectangle, then its area is​

Detailed Solution for Test: Distance Formula - Question 5

Let A(0-1), B(6,7), C(-2,3) and D(8,3) be the given points. Then

∴ AD = BC and AC = BD

So, ADBC is a parallelogram

Now 

Clearly, AB= AD+ DBand CD= CB+ BD2

Hence, ADBC is a rectangle.

Now

,Area of rectangle ADBC = AD × DB =(4√5​ × 2√5​)sq. units = 40sq. units

Test: Distance Formula - Question 6

The perimeter of a triangle with vertices (0, 4) (0, 0) and (3, 0) is:

Detailed Solution for Test: Distance Formula - Question 6

 

Test: Distance Formula - Question 7

The ordinate of a point is twice its abscissa. If its distance from the point (4,3) is √10,  then the coordinates of the point are​

Test: Distance Formula - Question 8

The perimeter of the triangle formed by the points A(0,0), B(1,0) and C(0,1) is

Test: Distance Formula - Question 9

The value of k, if the point P(0,2) is equidistant from A(3,k) and B(k,5) is

Detailed Solution for Test: Distance Formula - Question 9

 

Test: Distance Formula - Question 10

If A and B are the points (-6, 7) and (-1, -5) respectively, then the distance 2AB is equal to​

Test: Distance Formula - Question 11

The values of x and y, if the distance of the point (x,y) from (-3,0) as well as from (3,0) is 4 are

Test: Distance Formula - Question 12

The point (-1,-5) lies in the Quadrant​

Test: Distance Formula - Question 13

The distance between the points (3,4) and (8,-6) is​

Detailed Solution for Test: Distance Formula - Question 13

Distance formula= 

Test: Distance Formula - Question 14

The point on y-axis that is equidistant from (2,3) and (-4,1) is​

Detailed Solution for Test: Distance Formula - Question 14

A(2, 3) and B(-4, 1) are the given points.

Let C(0.y) be the points are y - axis


Therefore, The point on y - axis is (0,-1)

Test: Distance Formula - Question 15

The point on x-axis which is equidistant from (5,9) and (-4,6) is​

Detailed Solution for Test: Distance Formula - Question 15

Any point on x-axis is of the form, (x,0)
It is equidistant from (5,9) and (-4,6)
Distance of (x,0) from (5,9)
=​​​​​​√(5 − x)2 + 92 ---- (i)

Distance of (x,0) from (-4,6)
= √(x + 4)2 + 62 ----- (ii)
(i) = (ii)
√(5 − x)2 + 92 = √(x + 4)2 + 62 ​​​​​​​
25 + x2−10x+81 = x2 + 16 + 8x + 36
106−52=18x
54=18x
x=3
(3,0) is the equidistant point.

Test: Distance Formula - Question 16

The distance of the point P(6,-6) from the origin is equal to​

Test: Distance Formula - Question 17

The distance between the points P (-6,7) and Q (-1,-5) is​

Test: Distance Formula - Question 18

The condition that the point (x,y) may lie on the line joining (3,4) and (-5,-6) is​

Detailed Solution for Test: Distance Formula - Question 18

Since the point P(x,y) lies on the line joining A(3,4) and B(-5,-6), 

Therefore, points P, A and B are collinear points.

So, area of triangle PAB = 0                                         

Therefore, we have: 

10x-18-3y-5y+20=0

10x-8y+2=0

5x-4y+1=0 , which is the required condition. 

Test: Distance Formula - Question 19

The points (3, 2), (0, 5), (-3, 2) and (0, -1) are the vertices of a quadrilateral. Which quadrilateral is it?​

Test: Distance Formula - Question 20

The graph of the equation x = 3 is:​

Detailed Solution for Test: Distance Formula - Question 20

x=3 is fixed. This means the value of x is constant. So y can vary but x has only one value. For example (3,0),(3,2),(3,5) etc. So the line drawn will be parallel to y axis as y can vary.

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