Question

1. triangle jkl with vertices j(-6, 4), k(-4, 2), and l(-6, -3): 90° counterclockwise about (-2, 5)

j(__, __)
k(__, __)
l(__, __)

Explanation:

Step1: Translate points relative to center

To rotate a point $(x,y)$ counter - clockwise $90^{\circ}$ about a center $(a,b)$, first translate the point so that the center is at the origin. The translation formula is $(x - a,y - b)$.
For point $J(-6,4)$ with center $(-2,5)$:
$x_J'=-6-(-2)=-4$
$y_J'=4 - 5=-1$
For point $K(-4,2)$ with center $(-2,5)$:
$x_K'=-4-(-2)=-2$
$y_K'=2 - 5=-3$
For point $L(-6,-3)$ with center $(-2,5)$:
$x_L'=-6-(-2)=-4$
$y_L'=-3 - 5=-8$

Step2: Apply 90 - degree counter - clockwise rotation formula

The formula for a $90^{\circ}$ counter - clockwise rotation about the origin of a point $(x,y)$ is $(-y,x)$.
For the translated point $J'(-4,-1)$:
$x_{J''}=-(-1)=1$
$y_{J''}=-4$
For the translated point $K'(-2,-3)$:
$x_{K''}=-(-3)=3$
$y_{K''}=-2$
For the translated point $L'(-4,-8)$:
$x_{L''}=-(-8)=8$
$y_{L''}=-4$

Step3: Translate back

To get the final points, use the reverse - translation formula $(x + a,y + b)$.
For the rotated point $J''(1,-4)$:
$x_{J'}=1+(-2)=-1$
$y_{J'}= - 4+5 = 1$
For the rotated point $K''(3,-2)$:
$x_{K'}=3+(-2)=1$
$y_{K'}=-2 + 5=3$
For the rotated point $L''(8,-4)$:
$x_{L'}=8+(-2)=6$
$y_{L'}=-4 + 5=1$

Answer:

$J'(-1,1)$
$K'(1,3)$
$L'(6,1)$