Given, A=(2,3),B(3,−5),C=(x,y) Let centroid of △ABC is (h,k) ∴h=
2+3+x
3
,k=
3−5+y
3
⇒x=3h−5⇒y=3k+2 C lie on line 3x+4y−5=0 ∴3(3h−5)+4(3k+2)−5=0 ⇒9h−15+12k+8−5=0 ⇒9h+12k−12=0⇒3h+4k−4=0 Locus of centroid is 3x+4y−4=0 Which is parallel to line L=0