Distance Formula - Free MCQ Practice Test with solutions, Class 10

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.

We start with the fact that point P(x, y) is equidistant from A(7, 1) and B(3, 5). Using the distance formula:

  sqrt((x - 7)² + (y - 1)²) = sqrt((x - 3)² + (y - 5)²)

Square both sides to remove the square roots:

  (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²

Expand both sides:

  Left side: x² - 14x + 49 + y² - 2y + 1
  Right side: x² - 6x + 9 + y² - 10y + 25

Simplify each side:

  Left side becomes: x² + y² - 14x - 2y + 50
  Right side becomes: x² + y² - 6x - 10y + 34

Subtract the right side from the left side:

  (x² + y² - 14x - 2y + 50) - (x² + y² - 6x - 10y + 34) = 0

This simplifies to:

  -8x + 8y + 16 = 0

Divide the entire equation by 8:

  -x + y + 2 = 0

Rearrange to solve for the relation between x and y:

  x - y = 2

Thus, the correct relation is x - y = 2, which corresponds to option C.

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, AB² = AD² + DB² and CD² = CB² + BD²

Hence, ADBC is a rectangle.

Now

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

Distance formula= 

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)

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.

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. 

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.