Question

1. the image of ( a(-3,-5) ) under a translation is ( a^{prime}(6,-1) ). find the image of ( b(3,-2) ) under the same translation.
2. explain why ( \triangle a^{prime} b^{prime} c^{prime} ) with vertices ( a^{prime}(-1,-2), b^{prime}(0,0) ), and ( c^{prime}(-6,0) ) is not a translation image of ( \triangle a b c ) with vertices ( a(1,2), b(0,0) ), and ( c(6,0) ). hint: graph it!
3. determine whether ( \triangle p^{prime} q^{prime} r^{prime} ) is a translation image of ( \triangle p q r ). explain.
4. determine the translation vector that moves every point of a preimage 4 units left and 6 units up.

Explanation:

Step1: Find the translation rule

The translation rule is \((x,y)\to(x + a,y + b)\). For point \(A(-3,-5)\) and its image \(A'(6,-1)\), we have \(-3+a = 6\) and \(-5 + b=-1\).
Solving \(-3+a = 6\) gives \(a=6+3 = 9\). Solving \(-5 + b=-1\) gives \(b=-1 + 5=4\). So the translation rule is \((x,y)\to(x + 9,y + 4)\).

Step2: Apply the translation rule to point \(B\)

For point \(B(3,-2)\), using the rule \((x,y)\to(x + 9,y + 4)\), we substitute \(x = 3\) and \(y=-2\).
The \(x\)-coordinate of the image is \(3+9=12\), and the \(y\)-coordinate of the image is \(-2 + 4=2\).

Answer:

The image of \(B(3,-2)\) is \(B'(12,2)\)