# Finding vertices of rhombus formed by lines $y=2x+4$, $y=-\frac{1}{3}x+4$ and $(12,0)$ is a vertex. Can't find last vertex.

The equations of two adjacent sides of a rhombus are $y=2x+4$, $y=-\frac{1}{3}x+4$. If $(12,0)$ is one vertex and all vertices have positive coordinates, find the coordinates of the other three vertices. (Need help finding last vertex)

I know that one of the vertices is $(4,0)$, from the fact that the two lines in the question intersect at that point.

I then found the length of the line between $(12,0)$ and $(0,4)$. $l=\sqrt{16+144}=\sqrt{160}$

A rhombus has 4 equal sides, and the opposite sides are parallel, so I know the length of all the sides and their gradients.

Let other two vertices be $P$ and $Q$, which are on the lines $y=2x+4$ and $y=2x-24$ respectively.

Length of line between $(0,4)$ and $P$: $\sqrt{(x-0)^2+(y-4)^2}=\sqrt{160}\Rightarrow \sqrt{x^2+(2x)^2}=\sqrt{160}\Rightarrow 5x^2=160\Rightarrow x=\pm4\sqrt{2}$

$x=4\sqrt{2}$ because question says all vertices have positive coordinates.

$y=2.4\sqrt2+4\Rightarrow y=4+8\sqrt{2}$, $P(4\sqrt{2},4+8\sqrt{2})$

So far, my answers match with the solutions provided in the book. When I try to use the same method to find $Q$ I get a solution that doesn't match, and I don't know why.

Length of line between $(12,0)$ and $Q$: $\sqrt{(x-12)^2+(y-0)^2}=\sqrt{160}\Rightarrow \sqrt{(x-12)^2+(2x-24)^2}=\sqrt{160}$
$\Rightarrow x^2-24x+144+4x^2-96x+576=160\Rightarrow x^2-24x+112=0 \Rightarrow x=12\pm4\sqrt{2}$, $x=12+4\sqrt{2}$, because this value of $x$ gives a +ve value for both coordinates, $y=8\sqrt{2}$.
Vertices are $(0,4),P(4\sqrt{2},4+8\sqrt{2}),Q(12+4\sqrt{2},8\sqrt{ 2}),(0,12)$.