Question

The diagonals of quadrilateral ABCD intersect at E (2,5). ABCD has vertices at A (3,7) and B (3,6). What must be the coordinates of Upper C and Upper D to ensure that ABCD is a​ parallelogram?

Answer 1

Answer:

C(1,3) and D(1,4).

Step-by-step explanation:

The given quadrilateral ABCD has vertices at A (3,7) and B (3,6). The diagonals of  this quadrilateral ABCD intersect at E (2,5).

Recall that, the diagonals of a parallelogram bisects each other.

This means that; E(2,5) is the midpoint of each diagonal.

Let C and D have coordinates C(m,n) and D(s,t)

Using the midpoint rule:

$(\frac{x_2+x_1}{2}, \frac{y_2+y_1}{2})$

The midpoint of AC is $(\frac{m+3}{2}, \frac{n+7}{2})=(2,5)$

This implies that;

$(\frac{m+3}{2}=2, \frac{n+7}{2}=5)$

$(m+3=4, n+7=10)$

$(m=4-3, n=10-7)$

$(m=1, n=3)$

The midpoint of BD is $(\frac{m+3}{2}, \frac{n+7}{2})=(2,5)$

This implies that;

$(\frac{s+3}{2}=2, \frac{t+6}{2}=5)$

$(s+3=4, t+6=10)$

$(s=4-3, t=10-6)$

$(s=1, t=4)$

Therefore the coordinates of C are (1,3) and D(1,4).