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 $PA = PB$
$PA = PB$
$PA^2 = PB^2$
$(x + 2)^2 + (0 - 5)^2 = (x - 2)^2 + (0 + 3)^2$
$\Rightarrow x^2 + 4 + 4x + 25 = x^2 + 4 - 4x + 9$
$\Rightarrow x^2 + 4x - x^2 + 4x = 4 + 9 - 4 - 25$
$\Rightarrow 8x = -16$
$\Rightarrow x = -2$
$\therefore$ The point on x-axis is $(-2, 0).$
Let $(x, 0)$ be the point on x-axis.
Such that $PA = PB$
$PA = PB$
$PA^2 = PB^2$
$(x + 2)^2 + (0 - 5)^2 = (x - 2)^2 + (0 + 3)^2$
$\Rightarrow x^2 + 4 + 4x + 25 = x^2 + 4 - 4x + 9$
$\Rightarrow x^2 + 4x - x^2 + 4x = 4 + 9 - 4 - 25$
$\Rightarrow 8x = -16$
$\Rightarrow x = -2$
$\therefore$ The point on x-axis is $(-2, 0).$
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