Lines through the sides of a triangle

Text

Given the triangle with vertices $A(3;-1)$, $B(2;-8)$, $C(0;5)$:

• (a) find the equation of line $AB$;
• (b) find the equation of line $BC$.

Solution

(a) Slope $m_{AB}=\dfrac{-8-(-1)}{2-3}=\dfrac{-7}{-1}=7$; through $A(3;-1)$: $y+1=7(x-3)\Rightarrow y=7x-22$, i.e. $7x-y-22=0$.

(b) Slope $m_{BC}=\dfrac{5-(-8)}{0-2}=\dfrac{13}{-2}=-\tfrac{13}{2}$; through $C(0;5)$: $y=-\tfrac{13}{2}x+5$, i.e. $13x+2y-10=0$.

$\boxed{AB:\ 7x-y-22=0,\quad BC:\ 13x+2y-10=0}$