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.

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

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