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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.

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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 $\left(\frac85,\frac{16}5\right)$.

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