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

Answer:

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

Step-by-step 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.