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Sides of a rhombus are parallel to the lines $x+y-1=0$ and $7x-y-5=0$. The diagonals of the rhombus intersect at $(1,3)$ and one of its vertices $A$ lies on the line $y=2x$. Find coordinates of vertex $A$.

My attempt:

We are given the slopes of the sides of rhombus. Then I assumed the lines as $y=7x+k$ and the other as $y=-x+j$. I could find the point of intersection, then it all became very absurd. Can you please suggest me any other more efficient and quick way? Geometrical answers are welcomed too.

Parcly Taxel
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user537610
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1 Answers1

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Let $a$ be the angle made by the sides parallel to $x+y-1=0$ with the $+x$-axis and $b$ the corresponding angle for the sides parallel to $7x-y-5=0$. Clearly, $\tan a=-1$ and $\tan b=7$. We want the possible values of $\tan\frac{a+b}2$, the slopes of the diagonals, since they bisect the rhombus's angles and bisect themselves perpendicularly.

$\tan(a+b)$ is easy to find using angle addition: $\frac{-1+7}{1-(-1\cdot7)}=\frac34$. The same identity then allows us to find $\tan\frac{a+b}2=x$: $$\frac34=\frac{2x}{1-x^2}$$ $$3x^2+8x-3=0$$ $$x=\frac13\lor x=-3$$ The diagonals of the rhombus have slopes $\frac13$ and $-3$. They pass through $(1,3)$, so are parts of the lines $y=\frac13x+\frac83$ and $y=-3x+6$. The possible coordinates of $A$ are thus the intersections of these lines with $y=2x$: $\left(\frac65,\frac{12}5\right)$ and $\left(\frac85,\frac{16}5\right)$.

Parcly Taxel
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