1. ## Rumbus

Hi there, I have a math problem that I am having bit of a problem with. The question said that the equation of two adjacent sides of a rhombus are $y= 2x +4$, $y= -\frac{1}{3}+4$. If (12,0) is one vertex and all vertices have positive coordinates find the coordinates of the other three vertices. Can someone please help me with this problem I am trying to get the answer in radix. thank you in advance.

2. Hello, scrible!

The equation of two adjacent sides of a rhombus are $y\,=\, 2x +4,\;y\,=\,-\tfrac{1}{3}+4$.
If (12,0) is one vertex and all vertices have positive coordinates,
find the coordinates of the other three vertices.
We are given quite a lot of information . . .
Code:
        |     C
|     o
|    *    *
|   *         *
|  *              o D
| *              *
B |*              *
(0,4)o              *
|   *         *
|       *    *
- - + - - - - - o - - - - - - -
|        (12,0)
|           A

We are given vertex $A(12,0).$

Both lines have a $y$-intercept of 4.
. . So another vertex is: . $\boxed{B(0,4)}$

Side $AB$ has length: . $\sqrt{(0-12)^2 + (4-0)^2} \:=\:\sqrt{160} \:=\:4\sqrt{10}$

Point $C$ is on the line $y \:=\:2x+4$
Point $C$ has coordinates: . $C(x,\:2x+4)$

Since side $BC$ = side $AB$, we have: . $\sqrt{(x-0)^2 + (2x+4-4)^2} \:=\:\sqrt{160}$

. . $\sqrt{x^2 + 4x^2} \:=\:\sqrt{160}\quad\Rightarrow\quad \sqrt{5x^2} \:=\:\sqrt{160} \quad\Rightarrow\quad x^2 \:=\:32 \quad\Rightarrow\quad x \:=\:4\sqrt{2}$

Then: . $y \:=\:2(4\sqrt{2}) + 4 \quad\Rightarrow\quad y \:=\:8\sqrt{2} + 4$

Therefore: . $\boxed{C\left(4\sqrt{2},\;8\sqrt{2}+4\right) }$

Since $AD \parallel BC\,\text{ and }\,AD \,=\, BC$, we can use this trick.

Going from $B$ to $C$, we moved: . $\text{right }4\sqrt{2}\:\text{ and }\,\text{ up }8\sqrt{2}$

Going from $A(12,0)$ to $D$, we will make the same moves.

. . And we arrive at: . $\boxed{D\left(12+4\sqrt{2},\;8\sqrt{2}\right)}$