Test: Distance Formula - Class 10 MCQ

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.

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.