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# The three vertices of a parallelogram $ABCD$ are $A(3,-4),B(-1,-3)$ and $C(-6,2)$. Find the coordinates of vertex $D$ and find the area of $ABCD$.

Last updated date: 20th Jun 2024
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Hint: We are given the three vertices of a parallelogram $ABCD$ are $A(3,-4),B(-1,-3)$ and $C(-6,2)$. By using midpoint $O$ find vertex $D$. After that, to find the area used, Area of parallelogram $ABCD$ $=$ Area of $\Delta ABC$ $+$ Area $\Delta ADC$.

Now we are given the three vertices of a parallelogram $ABCD$ are $A(3,-4),B(-1,-3)$ and $C(-6,2)$.
$O$ is midpoint of $AC$$=(\dfrac{3-6}{2},\dfrac{-4+2}{2}) Simplifying we get, O is midpoint of AC$$=(\dfrac{-3}{2},-1)$ ……… (1)
$O$ is the midpoint of $BD$$=(\dfrac{-1+a}{2},\dfrac{-3+b}{2}) …………. (2) Now equating (1) and (2), we get, (\dfrac{-3}{2},-1)=(\dfrac{-1+a}{2},\dfrac{-3+b}{2}) So, -\dfrac{3}{2}=\dfrac{-1+a}{2} a=-2 Now, \dfrac{-3+b}{2}=-1 b=1 Therefore, D(-2,1). Now, Area of parallelogram ABCD = Area of \Delta ABC + Area \Delta ADC Area of parallelogram ABCD = 2Area of \Delta ABC Let us find Area of \Delta ABC , Area of \Delta ABC =\dfrac{1}{2}[{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})] Area of \Delta ABC $=\dfrac{1}{2}[3(1-2)-2(2+4)-6(-4-1)]$ Simplifying we get, Area of \Delta ABC $=\dfrac{15}{2}$sq. units So now, Area of parallelogram ABCD = 2Area of \Delta ABC$$=2\times \dfrac{15}{2}=15$ sq. units
Therefore, the area of parallelogram is $15$ square. units.