Find an equation of the line that satisfies the given conditions.
Let the line passing(-8,-8) be L1 and another one be L2
Slope of L2 : $${\frac{\left({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$=$${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$
Since L1 perpendicular to L2
Slope of L1 x Slope of L2 = -1
Slope of L1 x $${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$ = -1
Slope of L1 = 2
Equation of L1 : $${\frac{\left({\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}}$$=2
y+8=2x+16
2x-y+8=0
Therefore, equation of the line passing through (-8,-8) is 2x-y+8=0
Let the line passing(-8,-8) be L1 and another one be L2
Slope of L2 : $${\frac{\left({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$=$${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$
Since L1 perpendicular to L2
Slope of L1 x Slope of L2 = -1
Slope of L1 x $${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$ = -1
Slope of L1 = 2
Equation of L1 : $${\frac{\left({\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}}$$=2
y+8=2x+16
2x-y+8=0
Therefore, equation of the line passing through (-8,-8) is 2x-y+8=0