# Given point (-5,12) 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?

Sep 15, 2017

Distance of point from origin is $13$ unit and is at an angle of
${247.38}^{0}$ from ${0}^{0}$

#### Explanation:

Point is at $\left(- 5 , 12\right)$ and origin is at $\left(0 , 0\right)$ , We know

the distance between two points $\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$ is

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

$D = \sqrt{{\left(- 5 - 0\right)}^{2} + {\left(12 - 0\right)}^{2}} = \sqrt{169} = 13$ . The point is on

$3$rd quadrant . $\tan \alpha = \frac{12}{5} \therefore \alpha = {\tan}^{-} 1 \left(\frac{12}{5}\right)$ or

$\alpha = {67.38}^{0} \therefore \theta = 180 + 67.38 = {247.38}^{0}$

Distance of point from origin is $13$ unit and is at an angle of

${247.38}^{0}$ from ${0}^{0}$ [Ans]