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Question

# If three consecutive vertices of a parallelogram ABCD are A(1, −2), B(3, 6) and C(5, 10), find its fourth vertex D.

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Solution

## Let A(1, −2), B(3, 6) and C(5, 10) be the three vertices of a parallelogram ABCD and the fourth vertex be D(a, b). Join AC and BD intersecting at O. We know that the diagonals of a parallelogram bisect each other. Therefore, O is the midpoint of AC as well as BD. $\mathrm{Midpoint}\mathrm{of}AC=\left(\frac{1+5}{2},\frac{-2+10}{2}\right)=\left(\frac{6}{2},\frac{8}{2}\right)=\left(3,4\right)\phantom{\rule{0ex}{0ex}}\mathrm{Midpoint}\mathrm{of}BD=\left(\frac{3+a}{2},\frac{6+b}{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\frac{3+a}{2}=3\text{and}\frac{6+b}{2}=4\phantom{\rule{0ex}{0ex}}⇒3+a=6\text{and}6+b=8\phantom{\rule{0ex}{0ex}}⇒a=6-3\text{and}b=8-6\phantom{\rule{0ex}{0ex}}⇒a=3\text{and}b=2$ Therefore, the fourth vertex is D(3, 2).

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