Question

for exercises 6 and 7, triangle jkl has vertices j(3, 1), k(2, -1), and l(1, 2). write the coordinate notation for each rotation given. then write the coordinates of △jkl’ after each rotation. 6. clockwise rotation of 90° about the origin 7. counterclockwise rotation of 180° about the origin

Explanation:

Step1: Recall 90 - degree clock - wise rotation rule

For a point $(x,y)$ rotated 90 - degree clock - wise about the origin, the new coordinates $(x',y')$ are given by $(y,-x)$.
For point $J(3,1)$:
$x = 3$, $y = 1$, so $x'=1$, $y'=- 3$, new coordinates of $J$ are $J'(1,-3)$.
For point $K(2, - 1)$:
$x = 2$, $y=-1$, so $x'=-1$, $y'=-2$, new coordinates of $K$ are $K'(-1,-2)$.
For point $L(1,2)$:
$x = 1$, $y = 2$, so $x'=2$, $y'=-1$, new coordinates of $L$ are $L'(2,-1)$.

Step2: Recall 180 - degree counter - clockwise rotation rule

For a point $(x,y)$ rotated 180 - degree counter - clockwise (same as 180 - degree clock - wise) about the origin, the new coordinates $(x',y')$ are given by $(-x,-y)$.
For point $J(3,1)$:
$x = 3$, $y = 1$, so $x'=-3$, $y'=-1$, new coordinates of $J$ are $J''(-3,-1)$.
For point $K(2,-1)$:
$x = 2$, $y=-1$, so $x'=-2$, $y' = 1$, new coordinates of $K$ are $K''(-2,1)$.
For point $L(1,2)$:
$x = 1$, $y = 2$, so $x'=-1$, $y'=-2$, new coordinates of $L$ are $L''(-1,-2)$.

Answer:

1. $J'(1,-3)$, $K'(-1,-2)$, $L'(2,-1)$
2. $J''(-3,-1)$, $K''(-2,1)$, $L''(-1,-2)$