Let the equations of two adjacent sides of a parallelogram ABCD be 2x-3y=-23 and 5x+4y=23.

Question

Let the equations of two adjacent sides of a parallelogram $ABCD$ be $2x-3y=-23$ and $5x+4y=23$. If the equation of its one diagonal $AC$ is $3x+7y=23$ and the distance of $A$ from the other diagonal is $d$, then $50d^2$ is equal to \_\_\_\_\_\_\_.

Answer: 529

Accepted Answer

$A$ is the intersection of $2x-3y=-23$ and $3x+7y=23$: solving gives $A=(-4,5)$.

$C$ is the intersection of $5x+4y=23$ and $3x+7y=23$: solving gives $C=(3,2)$.

Midpoint of $AC$ (which is also the midpoint of $BD$, and where diagonal $BD$ passes through): $\left(-\dfrac12,\dfrac72\right)$.

Diagonal $BD$ has slope perpendicular-related to the parallelogram sides; equation works out to $7x+y=0$ (as computed in the source, using $B$ obtained from intersecting the two given adjacent sides at $(-1,7)$, and the diagonal through the intersection point of the two given lines and the midpoint of $AC$).

Distance of $A(-4,5)$ from $7x+y=0$:
$$d = \frac{|-28+5|}{\sqrt{50}} = \frac{23}{\sqrt{50}}$$

$$50d^2 = 50\times\frac{529}{50} = 529$$

Solutions · 1