The sides of a rhombus ABCD are parallel to the lines, x-y+2=0 and 7x-y+3=0.

Question

The sides of a rhombus $ABCD$ are parallel to the lines, $x-y+2=0$ and $7x-y+3=0$. If the diagonals of the rhombus intersect at $P(1,2)$ and the vertex $A$ (different from the origin) is on the $y$-axis, then the ordinate of $A$ is:

Accepted Answer

The diagonals of the rhombus are along the angle bisectors of the two given side-lines:
$$\frac{x-y+2}{\sqrt2}=\pm\frac{7x-y+3}{5\sqrt2} \implies 5(x-y+2)=\pm(7x-y+3)$$

This gives $2x+4y-7=0$ (slope $-\dfrac12$) and $12x-6y+13=0$ (slope $2$).

Let $A=(0,c)$; slope of $AP$ (through $A(0,c)$ and $P(1,2)$) $=2-c$, which must equal one of the diagonal slopes:

If $2-c=2 \implies c=0$ (rejected, this is the origin).

If $2-c=-\dfrac12 \implies c=\dfrac52$.

So the ordinate of $A$ is $\dfrac52$.

Solutions · 1