How do you find the magnitude of YZ given Y(-4,12) and Z(1,19)?

Sep 18, 2016

Use the distance formula to find $| \setminus \vec{Y Z} | = \sqrt{74} \approx 8.6$

Explanation:

Use the distance formula.

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

$\left({x}_{1} , {y}_{1}\right) = \left(- 4 , 12\right)$ and $\left({x}_{2} , {y}_{2}\right) = \left(1 , 19\right)$

|\vec(YZ) |= sqrt((1- -4)^2+(19-12)^2

$| \setminus \vec{Y Z} | = \sqrt{{5}^{2} + {7}^{2}}$

$| \setminus \vec{Y Z} | = \sqrt{25 + 49}$

$| \setminus \vec{Y Z} | = \sqrt{74} \approx 8.6$