If A(x1,y1),B(x2,y2)
and C(x3,y3) are the vertices of a triangle, then Area of Δ=21{x1(y2−y3)+x2(y3−y1) +x3(y1−y2)}
Given points are A(6,3),B(−3,5),C(4,−2) and P(x,y) ∴ΔABCΔPBC =2121[6(5+2)−3(−2−3)+4(3−5)x(5+2)−3(−2−y)+4(y−5)] =[42+15−87x+7y−14] =497x+7y−14=7x+y−2