How do you find all points that have an x -coordinate of –4 and whose distance from point (4, 2) is 10?

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Answer 1 · 2014-10-27T15:37:17

$D = root 2 ((x_1 - x_2)^2 + (y_1 - y_2)^2)$

We want $D = 10$

Let $(x_1, y_1)$ be $(4, 2)$
Let $(x_2, y_2)$ be $(-4, y_2)$

$10 = root 2 ((4 - (-4))^2 + (2 - y_2)^2)$

Squaring both sides we have

$(10 = root 2 ((4 - (-4))^2 + (2 - y_2)^2))^2$

$=> 100 = (4 - (-4))^2 + (2 - y_2)^2$

$=> 100 = (8^2) + (2 - y_2)^2$

$=> 100 = 64 + (4 - 4y_2 + (y_2)^2)$

$=> 36 = (y_2)^2 + 4y_2 + 4$

$=> (y_2)^2 + 4y_2 + 4 = 36$

$=> (y_2)^2 + 4y_2 - 32 = 0$

$=> (y_2 + 8)(y_2 - 4) = 0$

y_2 = -8, y_2 = 4