Question

# If $$A(1, 2), B(4, 3), C(6, 6)$$ are the vertices of a parallelogram $$ABCD,$$ then find the coordinate of the forth vertex.

Solution

## Let $$D\left(x,y\right)$$ be the coordinate of the fourth vertex  of the parallelogram $$ABCD$$We know that diagonals bisect each other in a parallelogram.$$\Rightarrow\,$$Mid-point of $$AC=$$Mid-point of $$BD$$Using mid-point formula,we have $$\left(x,y\right)=\left(\dfrac{{x}_{1}+{x}_{2}}{2},\dfrac{{y}_{1}+{y}_{2}}{2}\right)$$$$\Rightarrow\,\left(\dfrac{1+6}{2},\dfrac{2+6}{2}\right)=\left(\dfrac{x+4}{2},\dfrac{y+3}{2}\right)$$$$\Rightarrow\,\dfrac{x+4}{2}=\dfrac{7}{2},\,\dfrac{y+3}{2}=\dfrac{8}{2}$$$$\Rightarrow\,x+4=7,\,y+3=8$$$$\Rightarrow\,x=7-4=3,\,y=8-3=5$$$$\therefore\,$$ the fourth vertex is $$\left(3,5\right)$$Mathematics

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