91Ó°ÊÓ

The given vertices of the triangle are

$K\left(3,−1\right),A\left(2,6\right),T\left(5,1\right)\text{,}I\left(−4,1\right)\text{,}E\left(−3,−6\right)\text{and}S\left(−6,−1\right)$

We have to describe the relationship between triangle KAT and triangle IES.

We will find the lengths KA, KT and ATusing the distance formula.

$\begin{array}{l}KA=\sqrt{{\left(2−3\right)}^2+{\left(6−\left(−1\right)\right)}^2}\left[\text{since}K\left(3,−1\right)\text{and}A\left(2,6\right)\right]\\ KA=\sqrt{{\left(2−3\right)}^2+{\left(6+1\right)}^2}\\ KA=\sqrt{{\left(−1\right)}^2+{\left(7\right)}^2}\\ KA=\sqrt{1+49}\\ KA=\sqrt{50}\\ KA=5\sqrt{2}\end{array}$

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

$\begin{array}{l}AT=\sqrt{{\left(5−2\right)}^2+{\left(1−6\right)}^2}\left[\text{since}A\left(2,6\right)\text{and}T\left(5,1\right)\right]\\ AT=\sqrt{{\left(3\right)}^2+{\left(−5\right)}^2}\\ AT=\sqrt{9+25}\\ AT=\sqrt{34}\end{array}$

Then we find IE, IS and ESusing distance formula:

$\begin{array}{l}IE=\sqrt{{\left(−3−\left(−4\right)\right)}^2+{\left(−6−1\right)}^2}\left[\text{since}I\left(−4,1\right)\text{and}E\left(−3,−6\right)\right]\\ IE=\sqrt{{\left(−3+4\right)}^2+{\left(−6−1\right)}^2}\\ IE=\sqrt{{\left(1\right)}^2+{\left(−7\right)}^2}\\ IE=\sqrt{1+49}\\ IE=\sqrt{50}\\ IE=5\sqrt{2}\end{array}$

$\begin{array}{l}IS=\sqrt{{\left(−6−\left(−4\right)\right)}^2+{\left(−1−1\right)}^2}\left[\text{since}I\left(−4,1\right)\text{and}S\left(−6,−1\right)\right]\\ IS=\sqrt{{\left(−6+4\right)}^2+{\left(−1−1\right)}^2}\\ IS=\sqrt{{\left(−2\right)}^2+{\left(−2\right)}^2}\\ IS=\sqrt{4+4}\\ IS=\sqrt{8}\\ IS=2\sqrt{2}\end{array}$

$\begin{array}{l}ES=\sqrt{{\left(−6−\left(−3\right)\right)}^2+{\left(−1−\left(−6\right)\right)}^2}\left[\text{since}E\left(−3,−6\right)\text{and}S\left(−6,−1\right)\right]\\ ES=\sqrt{{\left(−6+3\right)}^2+{\left(−1+6\right)}^2}\\ ES=\sqrt{{\left(−3\right)}^2+{\left(5\right)}^2}\\ ES=\sqrt{9+25}\\ ES=\sqrt{34}\end{array}$

Then we find the ratio of the corresponding sides:

$\begin{array}{l}\frac{KA}{IE}=\frac{5\sqrt{2}}{5\sqrt{2}}=1\\ \frac{KT}{IS}=\frac{2\sqrt{2}}{2\sqrt{2}}=1\\ \frac{AT}{ES}=\frac{\sqrt{34}}{\sqrt{34}}=1\end{array}$

$\frac{KA}{IE}=\frac{KT}{IS}=\frac{AT}{ES}=1$

Hence, the relationship between triangle KAT and triangle IESis: they are similar.