# Finding coordinates of points using vectors? (See picture)

Nov 16, 2017

See below.

#### Explanation:

To find the coordinates of A B C, we need to find the intersection of the vector lines. This can be done equating each pair of vector lines.

To find point A we will use vector lines AB and AC. See diagram.

$A B = A C$

$\left(\begin{matrix}0 \\ 2\end{matrix}\right) + r \left(\begin{matrix}2 \\ 1\end{matrix}\right) = \left(\begin{matrix}0 \\ 5\end{matrix}\right) + t \left(\begin{matrix}1 \\ - 1\end{matrix}\right)$

$\textcolor{w h i t e}{\to} 2 r = t$
$2 + r = 5 - t$

Solving for $r$ and $t$:

$r = 3 - t$

$2 \left(3 - t\right) = t \implies t = 2$

$2 r = 2 \implies r = 1$

Using these in $A B = A C$

$\left(\begin{matrix}0 \\ 2\end{matrix}\right) + 1 \left(\begin{matrix}2 \\ 1\end{matrix}\right) = \left(\begin{matrix}0 \\ 5\end{matrix}\right) + 2 \left(\begin{matrix}1 \\ - 1\end{matrix}\right)$

$\left(\begin{matrix}2 \\ 3\end{matrix}\right) = \left(\begin{matrix}2 \\ 3\end{matrix}\right)$

Coordinate of A: $\textcolor{w h i t e}{8} \left(2 \textcolor{w h i t e}{1} , 3\right)$

To find point B we will use vector lines AB and BC.

$A B = B C$

$\left(\begin{matrix}0 \\ 2\end{matrix}\right) + r \left(\begin{matrix}2 \\ 1\end{matrix}\right) = \left(\begin{matrix}8 \\ 6\end{matrix}\right) + s \left(\begin{matrix}- 1 \\ - 2\end{matrix}\right)$

$\textcolor{w h i t e}{888} 2 r = 8 - s$
$2 + r = 6 - 2 s$

$r = 4 - 2 s$

$2 \left(4 - 2 s\right) = 8 - s \implies s = 0$
$2 + r = 6 - 2 \left(0\right) \implies r = 4$

Using these in $A B = B C$

$\left(\begin{matrix}0 \\ 2\end{matrix}\right) + 4 \left(\begin{matrix}2 \\ 1\end{matrix}\right) = \left(\begin{matrix}8 \\ 6\end{matrix}\right) + 0 \left(\begin{matrix}- 1 \\ - 2\end{matrix}\right)$

$\left(\begin{matrix}8 \\ 6\end{matrix}\right) = \left(\begin{matrix}8 \\ 6\end{matrix}\right)$

Coordinate of B: $\textcolor{w h i t e}{8} \left(8 \textcolor{w h i t e}{1} , 6\right)$

To find point C we will use vector lines AC and BC.

$A C = B C$

$\left(\begin{matrix}0 \\ 5\end{matrix}\right) + t \left(\begin{matrix}1 \\ - 1\end{matrix}\right) = \left(\begin{matrix}8 \\ 6\end{matrix}\right) + s \left(\begin{matrix}- 1 \\ - 2\end{matrix}\right)$

$\textcolor{w h i t e}{8888} t = 8 - s$
$5 - t = 6 - 2 s$

$5 - \left(8 - s\right) = 6 - 2 s \implies s = 3$
$t = 8 - 3 \implies t = 5$

Using these in $A C = B C$

$\left(\begin{matrix}0 \\ 5\end{matrix}\right) + 5 \left(\begin{matrix}1 \\ - 1\end{matrix}\right) = \left(\begin{matrix}8 \\ 6\end{matrix}\right) + 3 \left(\begin{matrix}- 1 \\ - 2\end{matrix}\right)$

$\left(\begin{matrix}5 \\ 0\end{matrix}\right) = \left(\begin{matrix}5 \\ 0\end{matrix}\right)$

Coordinate of C: $\textcolor{w h i t e}{8} \left(5 \textcolor{w h i t e}{1} , 0\right)$

To calculate the lengths of the sides, we could either use the distance formula or find vectors $\vec{A B} , \vec{A C} \mathmr{and} \vec{C B}$ and find their magnitude.

Using vectors.

$\vec{A B} = \left(\begin{matrix}8 \\ 6\end{matrix}\right) - \left(\begin{matrix}2 \\ 3\end{matrix}\right) = \left(\begin{matrix}6 \\ 3\end{matrix}\right)$

$\vec{A C} = \left(\begin{matrix}5 \\ 0\end{matrix}\right) - \left(\begin{matrix}2 \\ 3\end{matrix}\right) = \left(\begin{matrix}3 \\ - 3\end{matrix}\right)$

$\vec{C B} = \left(\begin{matrix}8 \\ 6\end{matrix}\right) - \left(\begin{matrix}5 \\ 0\end{matrix}\right) = \left(\begin{matrix}3 \\ 6\end{matrix}\right)$

$| | A B | | = \sqrt{\left({6}^{2}\right) + \left({3}^{2}\right)} = \sqrt{45} = 3 \sqrt{5}$

$| | A C | | = \sqrt{\left({3}^{2}\right) + \left(- {3}^{2}\right)} = \sqrt{18} = 3 \sqrt{2}$

$| | C B | | = \sqrt{\left({3}^{2}\right) + \left({6}^{2}\right)} = \sqrt{45} = 3 \sqrt{5}$

Sides AB and BC are of equal length.

Coordinates:

$A = \left(2 , 3\right)$
$B = \left(8 , 6\right)$
$C = \left(5 , 0\right)$

Side lengths:

$A B = 3 \sqrt{5}$
$B C = 3 \sqrt{5}$
$A C = 3 \sqrt{2}$