Sagot :
Answer:
Equation of line AB ,y−6=6–3−1–5(x+1)→x+2y=11
Equation of line BC ,y−3=3+25+5(x−5)→x−2y+1=0
Equation of line AB ,y+2=−2−6−5+1(x+1)→2x−y=0
Since, the equation of any line passing through (x1,y1) and (x2,y2) is
y−y1=y1−y2x1−x2(x−x1
Step-by-step explanation: