Angle between lines

Calculate the angle between these two lines:

$p:-8x+4y+5=0$

$q:10x+10y-7=0$

Correct result:

α =  45 °

Solution:

$\alpha =\arctan ({\displaystyle \frac{\mathrm{\mid }{\displaystyle \frac{-8}{4}}-{\displaystyle \frac{10}{10}}\mathrm{\mid }}{\mathrm{\mid }1-{\displaystyle \frac{-8}{4}}\cdot {\displaystyle \frac{10}{10}}\mathrm{\mid }}})=45^{\circ }$

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