Question

the coordinates of the vertices of the figure after the given transformation.
translation: $(x,y)\to(x - 4,y + 4)$
a $j(1,0),n(5,0),m(5,3),p(2,5)$
b $j(-2,5),n(2,5),m(2,2),p(-1,0)$
c $j(-3,4),n(1,4),m(1,1),p(-2,-1)$
d $j(-1,0),n(-5,0),m(-5,-3),p(-2,-5)$

Explanation:

Step1: Find the original coordinates

From the graph, the original coordinates are \(J(1,0)\), \(N(5,0)\), \(M(5, - 1)\), \(P(2,-5)\)

Step2: Apply the translation rule \((x,y)\to(x - 4,y + 4)\)

For point \(J\):
\(x=1,y = 0\)
\(x-4=1 - 4=-3\)
\(y + 4=0 + 4=4\)
New coordinate \(J'(-3,4)\)
For point \(N\):
\(x = 5,y=0\)
\(x-4=5 - 4=1\)
\(y + 4=0 + 4=4\)
New coordinate \(N'(1,4)\)
For point \(M\):
\(x = 5,y=-1\)
\(x-4=5 - 4=1\)
\(y + 4=-1 + 4=3\)
New coordinate \(M'(1,3)\) (There might be a mis - read in the original \(M\) coordinate from the graph. If we assume the original \(M\) is \((5,-1)\) and after translation \((1,3)\) is wrong. But if the original \(M\) is \((5,-3)\) (a more likely mis - read from the graph):
\(x = 5,y=-3\)
\(x-4=5 - 4=1\)
\(y + 4=-3 + 4=1\), so \(M'(1,1)\)
For point \(P\):
\(x = 2,y=-5\)
\(x-4=2 - 4=-2\)
\(y + 4=-5 + 4=-1\)
New coordinate \(P'(-2,-1)\)

Answer:

C. \(J(-3,4),N(1,4),M(1,1),P(-2,-1)\)