Answer :
[tex]Genera\ equation\ for\ line\\\\y=ax+b\\\\To\ find\ a\ and\ b\ substitude\ points\ (-2,3),\ (4,-1)\ into\ equation\\\\ \left \{ {{3=-2a+b}\ \ \ \ \atop {-1=4a+b\ |*-1}} \right.\\\\ \left \{ {{3=-2a+b}\ \ \ \ \atop {1=-4a-b\}} \right.\\+----\\\\Addition\ method\\\\4=-6a\ \ |:-6\\\\a=-\frac{4}{6}=-\frac{2}{3}\\\\b=3+2a=3+2* (-\frac{2}{3})=3-\frac{4}{3}=1\frac{2}{3}\\\\y=-\frac{2}{3}x+1\frac{2}{3}[/tex]
