Question

In Exercises 27-32 find and then compare lengths of segments.

Triangles JAN and RFK have vertices J(-2, -2), A(4, -2), M(2, 2), R(8, 1), F(S, 4), and K(6, 3). Show that triangle JAN is similar to triangle RFK.

Short answer

The triangle JAN and similar to the triangle RFK.

Step-by-step solution

Step-1 – Given

The given vertices of the triangle are

$J\left(-2,-2\right),A\left(4,-2\right),N\left(2,2\right)\text{,}R\left(8,1\right)\text{,}F\left(8,4\right)\text{and}K\left(6,3\right)$.

Step-2 – To determine

We have to show that the triangle JAN and similar to the triangle RFK.

Step-3 – Proof

We will first find JA, JN and ANusing the distance formula.

$\begin{array}{l}JA=\sqrt{{\left(4-\left(-2\right)\right)}^2+{\left(-2-\left(-2\right)\right)}^2}\left[\text{since}J\left(-2,-2\right)\text{and}A\left(4,-2\right)\right]\\ JA=\sqrt{{\left(4+2\right)}^2+{\left(-2+2\right)}^2}\\ JA=\sqrt{{\left(6\right)}^2+{\left(0\right)}^2}\\ JA=\sqrt{36}\\ JA=6\end{array}$

$\begin{array}{l}JN=\sqrt{{\left(2-\left(-2\right)\right)}^2+{\left(2-\left(-2\right)\right)}^2}\left[\text{since}J\left(-2,-2\right)\text{and}N\left(2,2\right)\right]\\ JN=\sqrt{{\left(2+2\right)}^2+{\left(2+2\right)}^2}\\ JN=\sqrt{{\left(4\right)}^2+{\left(4\right)}^2}\\ JN=\sqrt{16+16}\\ JN=\sqrt{32}\\ JN=4\sqrt{2}\end{array}$

$\begin{array}{l}AN=\sqrt{{\left(2-4\right)}^2+{\left(2-\left(-2\right)\right)}^2}\left[\text{since}A\left(4,-2\right)\text{and}N\left(2,2\right)\right]\\ AN=\sqrt{{\left(2-4\right)}^2+{\left(2+2\right)}^2}\\ AN=\sqrt{{\left(-2\right)}^2+{\left(4\right)}^2}\\ AN=\sqrt{4+16}\\ AN=\sqrt{20}\\ AN=2\sqrt{5}\end{array}$

Then, we find RF, RK and FKusing the distance formula.

$\begin{array}{l}RF=\sqrt{{\left(8-8\right)}^2+{\left(4-1\right)}^2}\left[\text{since}R\left(8,1\right)\text{and}F\left(8,4\right)\right]\\ RF=\sqrt{{\left(0\right)}^2+{\left(3\right)}^2}\\ RF=\sqrt{0+9}\\ RF=\sqrt{9}\\ RF=3\end{array}$

$\begin{array}{l}RK=\sqrt{{\left(6-8\right)}^2+{\left(3-1\right)}^2}\left[\text{since}R\left(8,1\right)\text{and}K\left(6,3\right)\right]\\ RK=\sqrt{{\left(-2\right)}^2+{\left(2\right)}^2}\\ RK=\sqrt{4+4}\\ RK=\sqrt{8}\\ RK=2\sqrt{2}\end{array}$

$\begin{array}{l}FK=\sqrt{{\left(6-8\right)}^2+{\left(3-4\right)}^2}\left[\text{since}F\left(8,4\right)\text{and}K\left(6,3\right)\right]\\ FK=\sqrt{{\left(-2\right)}^2+{\left(-1\right)}^2}\\ FK=\sqrt{4+1}\\ FK=\sqrt{5}\end{array}$

Then we check the ratio of the corresponding sides:

$\begin{array}{l}\frac{JA}{RF}=\frac{6}{3}=2\\ \frac{JN}{RK}=\frac{4\sqrt{2}}{2\sqrt{2}}=2\\ \frac{AN}{FK}=\frac{2\sqrt{5}}{\sqrt{5}}=2\end{array}$

So,

$\frac{JA}{RF}=\frac{JN}{RK}=\frac{AN}{FK}=2$

It means the triangle JAN and similar to the triangle RFK.