3.given $ \triangle j k l $ with vertices $ j ( - 2,8 ) , k ( 1 , - 3 ) $, and $ l ( 5,4 ) $ write the coordinates $ j ^ { prime } , k ^ { prime } $, and $ l ^ { prime } $ after the first translation, then $ j ^ { prime prime } , k ^ { prime prime } $, and $ l ^ { prime prime } $ after the second translation. translation: $ ( x , y ) \
ightarrow ( x + 6 , y - 1 ) $ translation: $ ( x , y ) \
ightarrow ( x - 1 , y - 7 ) $ $ j ( - 2,8 ) $ $ j ^ { prime } ( \quad , \quad ) $ $ j ^ { prime prime } ( \quad , \quad ) $ $ k ( 1 , - 3 ) $ $ k ^ { prime } ( \quad , \quad ) $ $ k ^ { prime prime } ( \quad , \quad ) $ $ l ( 5,4 ) $ $ l ^ { prime } ( \quad , \quad ) $ $ l ^ { prime prime } ( \quad , \quad ) $

Question

Accepted Answer

# Explanation:
## Step1: Find $J'$, $K'$, $L'$
For $J(-2,8)$:
$x=-2$, $y = 8$. Using $(x,y)\to(x + 6,y-1)$
$x'=-2+6 = 4$, $y'=8 - 1=7$. So $J'(4,7)$
For $K(1,-3)$:
$x = 1$, $y=-3$. Using $(x,y)\to(x + 6,y-1)$
$x'=1+6=7$, $y'=-3-1=-4$. So $K'(7,-4)$
For $L(5,4)$:
$x = 5$, $y = 4$. Using $(x,y)\to(x + 6,y-1)$
$x'=5+6 = 11$, $y'=4-1 = 3$. So $L'(11,3)$

## Step2: Find $J''$, $K''$, $L''$
For $J'(4,7)$:
$x = 4$, $y = 7$. Using $(x,y)\to(x-1,y - 7)$
$x''=4-1=3$, $y''=7-7 = 0$. So $J''(3,0)$
For $K'(7,-4)$:
$x = 7$, $y=-4$. Using $(x,y)\to(x-1,y - 7)$
$x''=7-1=6$, $y''=-4-7=-11$. So $K''(6,-11)$
For $L'(11,3)$:
$x = 11$, $y = 3$. Using $(x,y)\to(x-1,y - 7)$
$x''=11-1 = 10$, $y''=3-7=-4$. So $L''(10,-4)$

# Answer:
$J'(4,7)$, $K'(7,-4)$, $L'(11,3)$, $J''(3,0)$, $K''(6,-11)$, $L''(10,-4)$

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QUESTION IMAGE

Question

Explanation:

Step1: Find \(J'\), \(K'\), \(L'\)

Step2: Find \(J''\), \(K''\), \(L''\)

Answer: