# LINR 3 | Lesson 4 | Explore 3 Solution

Use systems of equations to describe a triangle defined in a coordinate plane.

Suppose our triangle looks like this:

A triangle will be a right triangle if the slopes of the legs are opposite reciprocals (m and $$\dfrac{-1}{m}$$).

So, to find the systems of equations defining the triangle above:

Given $$y=2x+6$$

A line perpendicular has $$m=\dfrac{-1}{2}$$

$$y=\dfrac{-1}{2}x+b$$

Then at $$(1,4)$$:

$$4=\dfrac{-1}{2}\normalsize 1+b$$

$$b=\dfrac{9}{2}$$

$$y=\dfrac{-1}{2}x+\dfrac{9}{2}$$

$$x=1$$ is a line connecting $$(1,4)$$ and $$y=2x+6$$

So, our system defining our triangle and its area are

$$y\geq 2x+6$$

$$x\geq 1$$

$$y\geq \dfrac{-1}{2} x + \dfrac{9}{2}$$

To identify vertices of a figure identified by linear inequalities, solve the linear system of equations algebraically (use the graph to estimate the vertices).

To find the vertex connecting the legs, we solve the system

$$y=2x+6$$

$$y= -\dfrac{1}{2}x + \dfrac{9}{2}$$

Using substitution, we find the vertex at $$\left(\dfrac{21}{5}, \dfrac{12}{5}\right)$$

So, our vertices are: $$(1,4),\hspace{2mm} (1,-4),\hspace{2mm} \left(\dfrac{21}{5}, \dfrac{12}{5}\right)$$

We can use the pythagorean theorem and the coordinates of the vertices to find the length of the sides of the figure.

\begin{align}|CT|&=8 \\\\ \left(1- \dfrac{21}{5}\right)^2 + \left(4- \dfrac{12}{5}\right)^2&=|CA|^2 \\\\ |CA|&=\dfrac{8}{\sqrt{5}} \approx 3.6 \\\\ |AT|^2&=64-\dfrac{64}{5} \\\\ |AT|&=\dfrac{16}{\sqrt{5}} \approx 7.2 \end{align}

Perimeter is distance around the figure, so for a triangle, we add up the the lengths of the sides.

$P_{cat}=|CA|+|AT|+|CT|=\dfrac{8}{\sqrt{5}}+\dfrac{16}{\sqrt{5}}+8, \hspace{2mm} etc.$

Area is space inside the figure measured in square units. For a triangle, we identify and find the lengths of the base and height to find the area.

$A_{cat}=\dfrac{1}{2}(|CA|\cdot|AT|), \hspace{2mm} etc.$