Question

In parallelogram ABCD, A(2,3), B(7,3) and D(4,7). Then the coordinates of C is

• A. (9,7)
• B. (7,9)
• C. (9,4)
• D. (4,9)

Answer 1

The correct option is A

(9,7)

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 = 4 - 2 = 2 units and so, BF = 2 units.

Also, ED = FC.

Since FC = 7 - 3 = 4 units, we have, ED = 4 units.

$x$-coordinate of C = 7 + BF = 7 + 2 = 9 units.

$y$-coordinate of C = 3 + FC = 3 + 4 = 7 units.

Hence C has the coordinates (9,7).