Question
Find a point on the x-axis which is equidistant from the points $(7, 6)$ and $(-3, 4).$
✓
Answer
Let $A (7,6)$ and $B (-3,4)$ be the given points.
Let $P(x, 0)$ be the point on $x$-axis such that $P A=P B$
$PA = PB$
$PA^2 = PB^2$
$(x - 7)^2 + (0 - 6)^2 = (x + 3)^2 + (0 - 4)^2$
$\Rightarrow x^2 + 49 - 14x + 36 = x^2 + 9 + 6x + 16$
$\Rightarrow x^2 - 14x - x^2 - 6x = 9 + 16 - 36 - 49$
$\Rightarrow -20x = -60$
$\Rightarrow x = 3$
$\therefore$ The point on x-axis is $(3, 0).$
Let $P(x, 0)$ be the point on $x$-axis such that $P A=P B$
$PA = PB$
$PA^2 = PB^2$
$(x - 7)^2 + (0 - 6)^2 = (x + 3)^2 + (0 - 4)^2$
$\Rightarrow x^2 + 49 - 14x + 36 = x^2 + 9 + 6x + 16$
$\Rightarrow x^2 - 14x - x^2 - 6x = 9 + 16 - 36 - 49$
$\Rightarrow -20x = -60$
$\Rightarrow x = 3$
$\therefore$ The point on x-axis is $(3, 0).$
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