Question

draw the preimage and image of each triangle under a translation along $\langle -4,1\
angle$.

1. triangle with coordinates:

$p(2,-1)$, $q(2,-3)$, $r(4,-3)$.
transformation rule: $(x,y)\to$

Explanation:

Step1: Apply translation rule

A translation along \(\langle - 4,1
angle\) means \(x\) - coordinate of pre - image \((x,y)\) is transformed to \(x-4\) and \(y\) - coordinate is transformed to \(y + 1\) in the image.
For point \(P(2,-1)\):
\(x=2,y=-1\)
Image \(x\) - coordinate: \(2-4=-2\)
Image \(y\) - coordinate: \(-1 + 1=0\)
So, image of \(P\) is \(P'(-2,0)\)

Step2: Apply translation rule to \(Q\)

For point \(Q(2,-3)\):
\(x = 2,y=-3\)
Image \(x\) - coordinate: \(2-4=-2\)
Image \(y\) - coordinate: \(-3 + 1=-2\)
So, image of \(Q\) is \(Q'(-2,-2)\)

Step3: Apply translation rule to \(R\)

For point \(R(4,-3)\):
\(x = 4,y=-3\)
Image \(x\) - coordinate: \(4-4=0\)
Image \(y\) - coordinate: \(-3 + 1=-2\)
So, image of \(R\) is \(R'(0,-2)\)

Answer:

Pre - image coordinates: \(P(2,-1)\), \(Q(2,-3)\), \(R(4,-3)\)
Image coordinates: \(P'(-2,0)\), \(Q'(-2,-2)\), \(R'(0,-2)\)
To draw:

1. Plot pre - image points \(P(2,-1)\), \(Q(2,-3)\), \(R(4,-3)\) on the coordinate plane. Connect them to form the pre - image triangle.
2. Plot image points \(P'(-2,0)\), \(Q'(-2,-2)\), \(R'(0,-2)\) on the coordinate plane. Connect them to form the image triangle.