Question

In parallelogram ABCD, A(0,-7), B(5,-7) and D(2,-2). Then the coordinates of C are

• A. (7,-2)
• B. (7,-3)
• C. (6,-2)
• D. (6,-3)

Answer 1

The correct option is A

(7,-2)

In the figure, consider $△AED$ and $△BFC$.

AD = BC, DE = CF and $\angle E$, $\angle F$ = ${90}^{\circ }$

Therefore $△AED$ and $△BFC$ are congruent to each other (by RHS congruence).

So, AE = BF

We see that AE = 2 - 0 = 2 units and so, BF = 2 units.

Also, ED = FC.

Since FC = -2 - (-7) = 5 units, we have, ED = 5 units.

$x$-coordinate of C = 5 + BF = 5 + 2 = 7

$y$-coordinate of C = -7 + FC = -7 + 5 = -2

Hence C has the coordinates (7,-2).