Question

Find the equation to the straight line joining the point of intersection of lines $x-2y+3=0$ and $2x-3y+4=0$ to the point $(4,-5)$.

Answer 1

$2a+3\left(4\right)-1=0=>2a+12-1=0=>2a=-11=>a=\frac{-11}{2}2(-2)+3b-1=0=>-4+3b-1=0=>3b=5=>b=\frac{5}{3}6x+6y-25=6\times \left(\frac{-11}{2}\right)+6\times \left(\frac{5}{3}\right)-25=-33+10-25=-48$

Equation of the line joining $(4,-5)$ and $(1,2)$

$\frac{x-4}{y+5}=\frac{1-4}{2+5}=>\frac{x-4}{y+5}=\frac{-3}{7}=>7x-28=-3y-15=>7x+3y+7=0$