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Find a point on the x-axis which is equidistant from the points A

Question:

Find a point on the x-axis which is equidistant from the points A (7, 6) and B(- 3, 4).

Solution:

Let the point on x-axis be P(x, 0).

Given: Point $P(x, 0)$ is equidistant from points $A(7,6)$ and $B(-3,4)$

i.e., distance of $P$ from $A=$ distance of $P$ from $B$

$\Rightarrow \sqrt{(x-7)^{2}+36}=\sqrt{(x+3)^{2}+16}$

Squaring both sides,

$\Rightarrow(x-7)^{2}+36=(x+3)^{2}+16$

$\Rightarrow x^{2}-14 x+49+36=x^{2}+6 x+9+16$

$\Rightarrow-20 x=-60$

$\Rightarrow x=3$

Therefore, the point on the x-axis is (3, 0).