Question

The coordinates of a point on the X-axis which lie on the perpendicular

bisector of the line segment joining (7,6) and (-3,4) are

A) (0,2)

B) (3,0)

C) (0,3)

D) (2,0)​

Answer 1

Answer:

B) (3,0)$\sqrt({x}-7)^{2} +(y-6) ^{2} = \sqrt({x} -(3)) + (Y-4)^{2}\\(x-7)^{2} + (y-6)^{2} = (x+3)62 + (y-4)^{2}\\x^{2} - 14x +49+y^{2} -12y+36= x^{2} +6x+9+y^{2}-8y+16\\-14x-6x-12y-8y+49+36-9-16=0\\-20x-20y+60= 0\\x-y-3=0\\x-y=3$

Step-by-step explanation:

Let P(x,y)  be any point on the perpendicular bisector of A,B. Then,PA=PB

On x-axis y is 0, so substituting y=0 we get x=3

Hence, the coordinates of point is (3,0)