AP ICET 2015 Paper

Question : 139 of 200

Marks: +1, -0

The point on x-axis which is equidistant from the points

(7,6),(-3,4)

Solution:

Given : A(7,6), B(–3, 4)

Let the point of the x-axis be C(x,o)
Then AC = BC

(x-7)^2+(-6)^2=(x+3)^2+(-4)^2

x^2+49-14x+36=x^2+9+6x+16

20x=60

⇒x=3

∴ The required point C is (3, 0)

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