# 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 $$\Large \frac{-1}{m}$$).

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

Given $$y=2x+6$$

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

$$y=\Large \frac{-1}{2}\normalsize x+b$$

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

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

$$b=\Large \frac{9}{2}$$

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

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

So,our system defining our triangle and its area are

$$y\geq 2x+6$$

$$x\geq 1$$

$$y\geq \Large \frac{-1}{2} \normalsize x + \Large \frac{9}{2}$$

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

$$y=2x+6$$

$$y=\Large -\frac{1}{2}x\normalsize + \Large \frac{9}{2}$$

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

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

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

$|CT|=8$

$\left(1- \frac{21}{5}\right)^2 + \left(4- \frac{12}{5}\right)^2=|CA|^2$

$|CA|=\frac{8}{\sqrt{5}} \approx 3.6$

$|AT|^2=64-\frac{64}{5}$

$|AT|=\frac{16}{\sqrt{5}} \approx 7.2$

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

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

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.

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