User:Lartoven/Sandbox 4

$$ \cos{c} = \frac{z}{2\sqrt{5}}$$

$$z = \cos{c} * 2\sqrt{5}$$

$$ \text{Distance formula for } (x_1, y_1) , (x_2 , y_2)\text{ : } \sqrt{ (x_1 - x_2)^2 + (y_1 - y_2)^2 } \,\!$$

$$ \text{Midpoint formula for } (x_1, y_1) , (x_2 , y_2)\text{ :  } \left ( \frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2} \right ) $$

$$ \text{Midpoint }z\,\,\text{of Point A }(-6, 0)\text{and Point B }(-2,-2)= \left ( \frac{-6+-2}{2}, \frac{0+-2}{2} \right ) = \left ( -4, -1 \right ) $$

$$z = \sqrt{(-4-x)^2 + (-1-y)^2}$$

$$\cos{c} * 2\sqrt{5} = \sqrt{(-4-x)^2 + (-1-y)^2}$$

$$\cos{30^\circ} * 2\sqrt{5} = \sqrt{ 16 + 8x + x^2 +1 + 2y + y^2}$$

$$\left ( 2\sqrt{5}*\cos{30^\circ} \right ) ^2 = x^2 + y^2 + 8x + 2y + 17$$

$$20 * \frac{3}{4} = 15 = x^2 + y^2 + 8x + 2y + 17$$

$$15 = x^2 + y^2 + 8x + 2y \,\!$$

$$ \begin{cases} 15 = x^2 + y^2 + 8x + 2y \\ y = 2x + 7 \\ \end{cases} $$ $$ x = -4 - \sqrt{3}\quad,\quad y = -1 - 2\sqrt{3} $$

$$ x = \sqrt{3} - 4\quad,\quad y = 2\sqrt{3} - 1 $$