Question

Three consecutive vertices of a parallelogram ABCD are A(10,-6), B(2, -6) and C(-4,-2), find the fourth vertex D.​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{Three consecutive verrtices of a parallelogram ABCD are}$

$\mathsf{A(10,-6),B(2,-6),C(-4-2)}$

$\underline{\textbf{To find:}}$

$\textsf{Fourth vertex D}$

$\underline{\textbf{Solution:}}$

$\textsf{Let the fourth vertex be D(x,y)}$

$\textsf{We know that,}$

$\boxed{\textbf{Diagonals of parallelogram bisect each other}}$

$\textsf{Mid point of AC=Mid point of BD}$

$\mathsf{\left(\dfrac{10+(-4)}{2},\dfrac{-6+(-2)}{2}\right)=\left(\dfrac{2+x}{2},\dfrac{-6+y}{2}\right)}$

$\mathsf{\left(\dfrac{6}{2},\dfrac{-8}{2}\right)=\left(\dfrac{2+x}{2},\dfrac{-6+y}{2}\right)}$

$\mathsf{\left(3,-4\right)=\left(\dfrac{2+x}{2},\dfrac{-6+y}{2}\right)}$

$\mathsf{\dfrac{2+x}{2}=3\,\implies\,2+x=6\,\implies\,x=4}$

$\mathsf{\dfrac{-6+y}{2}=-4\,\implies\,-6+y=-8\,\implies\,y=-2}$

$\therefore\textbf{Fourth vertex is (4,-2)}$

Answer 2

The above answer is correct