Question

A(-1,0),B(3,1) and D(-2,1) are the vertices of a parallelogram ABCD. Find the coordinates of the fourth vertex C.

Answer 1

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

$\textsf{Vertices of parallelogram ABCD are A(-1,0), B(3,1), D(-2,1)}$

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

$\textsf{Fourth vertex C of the parallelogram}$

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

$\underline{\textsf{Concept used:}}$

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

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

$\textsf{Since the diagonals of parallelogram bisect each other,}$

$\textsf{we have,}$

$\textbf{Midpoint of diagonal AC=Midpoint of diagonal BD}$

$\implies\mathsf{\left(\dfrac{-1+x}{2},\dfrac{0+y}{2}\right)=\left(\dfrac{3+(-2)}{2},\dfrac{1+1}{2}\right)}$

$\implies\mathsf{\left(\dfrac{-1+x}{2},\dfrac{y}{2}\right)=\left(\dfrac{1}{2},\dfrac{2}{2}\right)}$

$\implies\mathsf{\left(\dfrac{-1+x}{2},\dfrac{y}{2}\right)=\left(\dfrac{1}{2},1\right)}$

$\textsf{Equating corresponding co-ordinates on bothsides, we get}$

$\mathsf{\dfrac{-1+x}{2}=\dfrac{1}{2}\;\;\&\;\;\dfrac{y}{2}=1}$

$\mathsf{-1+x=1\;\;\&\;\;y=2}$

$\mathsf{x=2\;\;\&\;\;y=2}$

$\therefore\textbf{The fourth vertex is (2,2)}}$

$\underline{\textbf{Find more:}}$

If A(1,3) ,B(-1,2),C(x,4) and D(2,5) are vertices of parallelogram then write the value of x​

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