Answer:
[tex]\text{Third option}[/tex]
Step-by-step explanation:
[tex]\text{There are two methods to rotate a point by 90 degrees counterclockwise.}[/tex]
[tex]\bold{1.\ Using\ Formula:}\text{To rotate a point (x,y) counterclockwise by 90 degrees,}\\\text{you can use the following formula:}\\(x,y)\rightarrow(-y,x)[/tex]
[tex]\text{Let's apply this formula to the given points:}\\\text{i. For point A:}\\\text{A(1,-2)}\rightarrow\text{A'}(2,1)[/tex]
[tex]\text{ii. For point B:}\\\text{B (-2,-1)}\rightarrow \text{B'(1,-2)}[/tex]
[tex]\bold{2.\ Using\ rotation\ matrix:}\\\text{To rotate the point (x,y) by counterclockwise direction by 90 degrees, you can}\\\text{use the following matrix formula:}\\\begin{bmatrix}x' \\y'\end{bmatrix}=\begin{bmatrix}\text{cos}\theta & \text{-sin}\theta \\\text{sin}\theta & \text{cos}\theta\end{bmatrix}\begin{bmatrix}x \\y\end{bmatrix}[/tex]
[tex]\text{where, }\theta\text{ is the angle of rotation and }(x',y') \text{ are the resulting coordinates after}\\\text{the rotation.}[/tex]
[tex]\text{Here, }\theta=90^\text{o}\text{ (If the question had said clockwise, it would be negative)}\\[/tex]
[tex]\text{i. Let's apply this matrix formula to point A:}\\\text{In this case, }(x,y)=(1,-2).\\[/tex]
[tex]\therefore\begin{bmatrix}x' \\y'\end{bmatrix}=\begin{bmatrix}\text{cos}90^\text{o} & \text{-sin}90^\text{o} \\\text{sin}90^\text{o} & \text{cos}90^\text{o}\end{bmatrix}\begin{bmatrix}1 \\-2\end{bmatrix}=\begin{bmatrix}0 & -1 \\1 & 0\end{bmatrix}\begin{bmatrix}1 \\-2\end{bmatrix}=\begin{bmatrix}0(1)+(-1)(-2) \\1(1)+(0)(-2)\end{bmatrix}\\\text{or, }\begin{bmatrix}x' \\y'\end{bmatrix}=\begin{bmatrix}2 \\1\end{bmatrix}[/tex]
[tex]\text{Equating corresponding elements,}\\x'=2\ \text{and }y'=1[/tex]
[tex]\text{ii. Now let's apply the matrix formula to point B:}\\\text{In this case, }(x,y)=(-2,-1).[/tex]
[tex]\therefore\begin{bmatrix}x' \\y'\end{bmatrix}=\begin{bmatrix}\text{cos}90^\text{o} & \text{-sin}90^\text{o} \\\text{sin}90^\text{o} & \text{cos}90^\text{o}\end{bmatrix}\begin{bmatrix}-2 \\-1\end{bmatrix}=\begin{bmatrix}0 & -1 \\1 & 0\end{bmatrix}\begin{bmatrix}-2 \\-1\end{bmatrix}=\begin{bmatrix}0(-2)+(-1)(-1) \\1(-2)+(0)(-1)\end{bmatrix}\\\text{or, }\begin{bmatrix}x' \\y'\end{bmatrix}=\begin{bmatrix}1 \\-2\end{bmatrix}[/tex]
[tex]\text{Equating corresponding elements,}\\x'=1\ \text{and }y'=-2\\\therefore\ \text{The rotated points are A'(2,1) and B'(1,-2).}[/tex]
Please ignore if you see 'amp;' in my answer.
