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

Oct 27, 2014

By using the distance formula.

$D = \sqrt[2]{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$

We want $D = 10$

Let $\left({x}_{1} , {y}_{1}\right)$ be $\left(4 , 2\right)$
Let $\left({x}_{2} , {y}_{2}\right)$ be $\left(- 4 , {y}_{2}\right)$

$10 = \sqrt[2]{{\left(4 - \left(- 4\right)\right)}^{2} + {\left(2 - {y}_{2}\right)}^{2}}$

Squaring both sides we have

${\left(10 = \sqrt[2]{{\left(4 - \left(- 4\right)\right)}^{2} + {\left(2 - {y}_{2}\right)}^{2}}\right)}^{2}$

$\implies 100 = {\left(4 - \left(- 4\right)\right)}^{2} + {\left(2 - {y}_{2}\right)}^{2}$

$\implies 100 = \left({8}^{2}\right) + {\left(2 - {y}_{2}\right)}^{2}$

$\implies 100 = 64 + \left(4 - 4 {y}_{2} + {\left({y}_{2}\right)}^{2}\right)$

$\implies 36 = {\left({y}_{2}\right)}^{2} + 4 {y}_{2} + 4$

$\implies {\left({y}_{2}\right)}^{2} + 4 {y}_{2} + 4 = 36$

$\implies {\left({y}_{2}\right)}^{2} + 4 {y}_{2} - 32 = 0$

$\implies \left({y}_{2} + 8\right) \left({y}_{2} - 4\right) = 0$

y_2 = -8, y_2 = 4