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The diagonals of quadrilateral ABCD intersect at E(−2,4). ABCD has vertices at A(1,7) B(−3,5). What must be the coordinates of C and D to ensure that ABCD is a​ parallelogram?

User Will Evers
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1 Answer

3 votes

Answer:

The coordinates of C and D are (1, 3) and (-5, 1), respectivelly.

Explanation:

Since E is the midpoint of diagonals AD and BC (see attachment). That is:


AD = 2 \cdot AE


BC = 2\cdot BE

The vectorial distances of AE and BE are, respectively:


\overrightarrow{AE} = \vec E - \vec A


\overrightarrow {AE} = (-2,4) -(1,7)


\overrightarrow{AE} = (-2-1, 4-7)


\overrightarrow {AE} = (-3,-3)


\overrightarrow{BE} = \vec E - \vec B


\overrightarrow {BE} = (-2,4) - (-3,5)


\overrightarrow {BE} = (-2+3, 4-5)


\overrightarrow {BE} = (1,-1)

Now, the relative vectorial distances to C and D are now obtained:


\overrightarrow {AD} = 2\cdot \overrightarrow {AE}


\overrightarrow {AD} = 2 \cdot (-3,-3)


\overrightarrow{AD} = (-6, -6)


\overrightarrow {BC} = 2 \cdot \overrightarrow {BE}


\overrightarrow{BC} = 2 \cdot (1,-1)


\overrightarrow {BC} = (2,-2)

Lastly, the coordinates are found by the following vectorial equations:


\vec C = \vec B + \overrightarrow {BC}


\vec C = (-3,5) + (2,-2)


\vec C = (-3+2, 5 -2)


\vec C = (-1,3)


\vec D = \vec A + \overrightarrow {AD}


\vec D = (1,7) + (-6,-6)


\vec D = (1-6, 7 -6)


\vec D = (-5, 1)

The coordinates of C and D are (1, 3) and (-5, 1), respectivelly.

The diagonals of quadrilateral ABCD intersect at E(−2,4). ABCD has vertices at A(1,7) B-example-1
User Jpoppe
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