Let ABCD be a parallelogram such that the coordinates of its three vertices A, B, C are (1, 1), (3, 4) and (−2, 8), respectively. Then, the coordinates of the vertex D are

Question

Let ABCD be a parallelogram such that the coordinates of its three vertices A, B, C are (1, 1), (3, 4) and (&minus;2, 8), respectively. Then, the coordinates of the vertex D are

Accepted Answer

(-4, 5) — Easy In any parallelogram the diagonals bisect each other, so the two diagonals share the same midpoint. For $ABCD$ in order, the diagonals are $AC$ and $BD$, hence $A+C=B+D$ (coordinate-wise). Solve for $D$. 1 Identify the diagonals. Vertices in order $A,B,C,D$, so opposite pairs are $A\!-\!C$ and $B\!-\!D$. Equal midpoints give: $$A+C=B+D$$ 2 Solve coordinate-wise with $A(1,1)$, $B(3,4)$, $C(-2,8)$: $$\begin{aligned} &x_D=x_A+x_C-x_B=1+(-2)-3=-4\ &y_D=y_A+y_C-y_B=1+8-4=5 \end{aligned}$$ 3 So

CAT 2022 Slot 1 — QA Question 8

Answer & solution