Two sides of a parallelogram are along the lines, x+y=3 and x-y+3=0.

Question

Two sides of a parallelogram are along the lines, $x+y=3$ and $x-y+3=0$. If its diagonals intersect at $(2,4)$, then one of its vertex is:

Accepted Answer

The two given lines intersect at $P(0,3)$ (solving $x+y=3, x-y=-3$). Since diagonals bisect each other at $M(2,4)$, and $P$ is one vertex, the opposite vertex $R = 2M-P = (4,5)$.

Let $S(x_1,x_1+3)$ on $x-y+3=0$ and $Q(x_2,3-x_2)$ on $x+y=3$ be the other two vertices; since $M$ is also midpoint of $SQ$:
$$x_1+x_2=4, \qquad (x_1+3)+(3-x_2)=8$$

Solving: $x_1=3,\ x_2=1$. So vertex $S=(3,6)$.

Solutions · 1