Question

If the image of the point $(-1,2)$ in the line $4x+3y-p=0$ is the point $(3,5)$, then find the value of $p$.

Answer 1

Let $A=(-1,2);B=(3,5)$

Mid point of A$AB\equiv (\frac{-1+3}{2},\frac{2+5}{2})\equiv (1,\frac{7}{2})$

$(1,\frac{7}{2})$ will satisfy the equation of line

So, $4\left(1\right)+3\left(\frac{7}{2}\right)-p=0$

$p=4+\frac{21}{2}=\frac{29}{2}$