# Given point (-8,8) how do you find the distance of the point from the origin, then find the measure of the angle in standard position whose terminal side contains the point?

##### 1 Answer
Mar 19, 2018

Measure of the angle $\theta = {135}^{\circ}$ or ${\left(\frac{3 \pi}{4}\right)}^{c}$

#### Explanation:

Distance formula : $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

Given point $A \left(- 8 , 8\right)$ and Origin $O \left(0 , 0\right)$

$\vec{O A} = r = \sqrt{{\left(- 8\right)}^{2} + {8}^{2}} = \sqrt{128} = 8 \sqrt{2}$

$x = r \cos \theta$, $y = r \sin \theta$

$- 8 = 8 \sqrt{2} \cos \theta$

$\cos \theta = - \frac{\cancel{8}}{\cancel{8} \sqrt{2}} = - \frac{1}{\sqrt{2}}$ or theta =color(purple)( - pi/4 = (pi - pi/4) = (3pi)/4

Similarly, $8 \sqrt{2} \sin \theta = 8$

$\sin \theta = \frac{8}{8 \sqrt{2}} = \frac{1}{\sqrt{2}}$ or $\theta = \frac{\pi}{4}$ or = pi - pi/4) = (3pi)/4

Slope of the line $m = \tan \theta = - \frac{8}{8} = - 1$

$x$ $\left(-\right)$ and $y$ $\left(+\right)$ in II quadrant.

Hence $\theta = {\tan}^{-} 1 \left(- \frac{8}{8}\right) = {\tan}^{-} 1 - 1 = - 1 = \frac{3 \pi}{4}$ ( )