Question

4. Square JKLM with vertices J(1, -3), K(5,0),L(8,-4), and M(4, -7): 90° counterclockwiseI'K'(27L'(M' (1-

Answer 1

Given the condition 90 degrees counterclockwise, we use rule below:

When rotating a point 90 degrees counterclockwise about the origin our point A(x,y) becomes A'(-y,x). In other words, switch x and y and make y negative.

Hence, To get J';

$\begin{gathered} A(x,y)=A^{\prime}(-y,x) \\ J(x,y)=J^{\prime}(-y,x) \\ J(1,-3)=J^{\prime}(-(-3),1) \\ J^{\prime}=(3,1) \end{gathered}$

To get K':

$\begin{gathered} K(x,y)=K^{\prime}^{}(-y,x) \\ K(5,0)=K^{\prime}(-0,5) \\ K^{\prime}=(0,5) \end{gathered}$

To get L':

$\begin{gathered} L(x,y)=L^{\prime}^{}(-y,x) \\ L(8,-4)=L^{\prime}(-(-4),8) \\ L^{\prime}=(4,8) \end{gathered}$

To get M':

$\begin{gathered} M(x,y)=M^{\prime}^{}(-y,x) \\ M(4,-7)=M^{\prime}(-(-7),4) \\ M^{\prime}=(7,4) \end{gathered}$

Therefore, the values of J',K',L' and M' are respectively given below;

$\begin{gathered} J^{\prime}=(3,1) \\ K^{\prime}=(0,5) \\ L^{\prime}=(4,8) \\ M^{\prime}=(7,4) \end{gathered}$