Question

If the point (x, 3) is equidistant from (3, -2) and (7, 4), find x.

Answer 1

Answer on Question #53147 – Math – Analytic Geometry

If the point $(x, 3)$ is equidistant from $(3, -2)$ and $(7, 4)$, find $x$.

Solution

(7;4)

$(x;3)$

$(3;-2)$

$$L_1 = \sqrt{(x - 7)^2 + (3 - 4)^2}$$ $$L_1 = \sqrt{x^2 - 14x + 49 + 1}$$ $$L_1 = \sqrt{x^2 - 14x + 50}$$ $$L_2 = \sqrt{(3 - x)^2 + (-2 - 3)^2}$$ $$L_2 = \sqrt{x^2 - 6x + 9 + 25}$$ $$L_2 = \sqrt{x^2 - 6x + 34}$$ $$L_2 = L_1$$ $$\sqrt{x^2 - 6x + 34} = \sqrt{x^2 - 14x + 50}$$ $$x^2 - 6x + 34 = x^2 - 14x + 50$$ $$14x - 6x = 50 - 34$$ $$8x = 16$$ $$x = 2$$

Answer: $x=2$.

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