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

$D = \sqrt[2]{{(x_{1} - x_{2})}_{1}^{2} + {(y_{1} - y_{2})}_{1}^{2}}$ $D = root 2 ((x_1 - x_2)^2 + (y_1 - y_2)^2)$

We want $D = 10$ $D = 10$

Let $(x_{1}, y_{1})$ $(x_1, y_1)$ be $(4, 2)$ $(4, 2)$
Let $(x_{2}, y_{2})$ $(x_2, y_2)$ be $(- 4, y_{2})$ $(-4, y_2)$

$10 = \sqrt[2]{{(4 - (- 4))}^{2} + {(2 - y_{2})}_{2}^{2}}$ $10 = root 2 ((4 - (-4))^2 + (2 - y_2)^2)$

Squaring both sides we have

${(10 = \sqrt[2]{{(4 - (- 4))}^{2} + {(2 - y_{2})}_{2}^{2}})}^{2}$ $(10 = root 2 ((4 - (-4))^2 + (2 - y_2)^2))^2$

$\Rightarrow 100 = {(4 - (- 4))}^{2} + {(2 - y_{2})}_{2}^{2}$ $=> 100 = (4 - (-4))^2 + (2 - y_2)^2$

$\Rightarrow 100 = (8^{2}) + {(2 - y_{2})}_{2}^{2}$ $=> 100 = (8^2) + (2 - y_2)^2$

$\Rightarrow 100 = 64 + (4 - 4 y_{2} + {(y_{2})}_{2}^{2})$ $=> 100 = 64 + (4 - 4y_2 + (y_2)^2)$

$\Rightarrow 36 = {(y_{2})}_{2}^{2} + 4 y_{2} + 4$ $=> 36 = (y_2)^2 + 4y_2 + 4$

$\Rightarrow {(y_{2})}_{2}^{2} + 4 y_{2} + 4 = 36$ $=> (y_2)^2 + 4y_2 + 4 = 36$

$\Rightarrow {(y_{2})}_{2}^{2} + 4 y_{2} - 32 = 0$ $=> (y_2)^2 + 4y_2 - 32 = 0$

$\Rightarrow (y_{2} + 8) (y_{2} - 4) = 0$ $=> (y_2 + 8)(y_2 - 4) = 0$

y_2 = -8, y_2 = 4