Question

triangle abc was created by joining points a (-8, 2), b (-3, 8) and c (-2, 3) with line segments. triangle abc is reflected over the y - axis and then reflected over the x - axis to form the triangle where x, y, and z represent the lengths of the sides of the triangle.

Explanation:

Step1: Find the coordinates after reflections

When reflecting a point $(x,y)$ over the $y -$axis, the transformation is $(-x,y)$. Then reflecting over the $x -$axis, the transformation is $(-x,-y)$.
For point $A(-8,2)$: After reflection over $y -$axis, it becomes $(8,2)$, then after reflection over $x -$axis, it becomes $(8,-2)$.
For point $B(-3,8)$: After reflection over $y -$axis, it becomes $(3,8)$, then after reflection over $x -$axis, it becomes $(3,-8)$.
For point $C(-2,3)$: After reflection over $y -$axis, it becomes $(2,3)$, then after reflection over $x -$axis, it becomes $(2,-3)$.

Step2: Use the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$ to find side - lengths

Let's find the length of the side between two points.
For the side between $(8,-2)$ and $(3,-8)$:
\[

$$\begin{align*} d&=\sqrt{(8 - 3)^2+(-2+8)^2}\\ &=\sqrt{5^2 + 6^2}\\ &=\sqrt{25+36}\\ &=\sqrt{61} \end{align*}$$

\]
For the side between $(3,-8)$ and $(2,-3)$:
\[

$$\begin{align*} d&=\sqrt{(3 - 2)^2+(-8 + 3)^2}\\ &=\sqrt{1^2+(-5)^2}\\ &=\sqrt{1 + 25}\\ &=\sqrt{26} \end{align*}$$

\]
For the side between $(2,-3)$ and $(8,-2)$:
\[

$$\begin{align*} d&=\sqrt{(8 - 2)^2+(-2 + 3)^2}\\ &=\sqrt{6^2+1^2}\\ &=\sqrt{36 + 1}\\ &=\sqrt{37} \end{align*}$$

\]
(The order of which side is $x$, $y$, $z$ is not specified in the problem - but these are the lengths of the sides of the new triangle).

Answer:

The side - lengths of the new triangle are $\sqrt{26}$, $\sqrt{37}$, $\sqrt{61}$