Question
Find the point on x-axis which is equidistant from the points $(-2, 5)$ and $(2, -3).$
✓
Answer
Let $A(-2, 5)$ and $(2, -3)$ be the given points.
Let $(x, 0)$ be the point on x-axis.
Such that $PA = PB$
$P A=P B$
$P A^2=P B^2$
$(x+2)^2+(0-5)^2=(x-2)^2+(0+3)^2$
$\Rightarrow x^2+4+4 x+25=x^2+4-4 x+9$
$\Rightarrow x^2+4 x-x^2+4 x=4+9-4-25$
$\Rightarrow 8 x=-16$
$\Rightarrow x=-2$
$\therefore$ The point on x-axis is $(-2, 0).$
Let $(x, 0)$ be the point on x-axis.
Such that $PA = PB$
$P A=P B$
$P A^2=P B^2$
$(x+2)^2+(0-5)^2=(x-2)^2+(0+3)^2$
$\Rightarrow x^2+4+4 x+25=x^2+4-4 x+9$
$\Rightarrow x^2+4 x-x^2+4 x=4+9-4-25$
$\Rightarrow 8 x=-16$
$\Rightarrow x=-2$
$\therefore$ The point on x-axis is $(-2, 0).$
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