Answer with explanation:
(a)→Coordinates of Vertices of triangle = (-2,-3), (3,5) and (8, -1).
Plotting these points in two dimensional coordinate plane:
Now when the points are reflected through X axis, the distance of points from line , y=0, on both sides is same .That is distance of Preimage from ,line, y=0 is same as Distance of Image from the line.
≡Rule of reflection of (x,y)→ (x, -y).
⇒Coordinates of Image =(-2,3), (3, -5), (8,1).
In Matrix form
![\left[\begin{array}{ccc}-2&-3\\3&5\\8&-1\end{array}\right] \text {reflected across X axis}=\left[\begin{array}{ccc}-2&3\\3&-5\\8&1\end{array}\right]](https://img.qammunity.org/2018/formulas/mathematics/high-school/bzupepikrkgrwcqrvhzqz1jieqw730853f.png)
(b) → Coordinates of Vertices of triangle = (-2,-3), (3,5) and (8, -1) is rotated by 90 degree, the vertices of triangle will be , (-3, 2), (5, -3), (-1, -8).
The rule is , after 90 degree rotation . point (a,b)→(b, -a).
Plotting these points in two dimensional coordinate plane:
In Matrix form
![\left[\begin{array}{ccc}-2&-3\\3&5\\8&-1\end{array}\right] \text {reflected across X axis}=\left[\begin{array}{ccc}-3&2\\5&-3\\-1&-8\end{array}\right]](https://img.qammunity.org/2018/formulas/mathematics/high-school/8i7r9w59mcperd1kacgs7gupa3fppe0fil.png)