Correct Answer - A
Solving the equations of the lines, we get
`A-=((1)/(1+m_(1)),(m_(1))/(1+m_(1))),C-=((1)/(1+m_(3)),(m_(3))/(1+m_(3)))`
If AB=BC, then the midpoint of AC lies on
`y=m_(2)x`
`"or "(m_(1)/(1+m1)+(m_(3))/(1+m3))/(2) = m_(2)[((1)/(1+m_(1))+(1)/(1+m_(3)))/(2)]`