Question

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?

Answer 1

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.