Question

triangle efg has vertices e(-3,4), f(-5,-1), and g(1,1). the triangle is translated so that the coordinates of the image are e(-1,0), f(-3,-5), and g(3,-3). which rule was used to translate the image? t_{4,-4}(x,y) t_{4,-4}(x,y) t_{2,-4}(x,y) t_{2,-4}(x,y)

Explanation:

Step1: Calculate the change in \(x\)-coordinate

For point \(E\): \(x\)-coordinate of \(E\) is \(-3\), \(x\)-coordinate of \(E'\) is \(-1\). The change \(\Delta x=-1 - (-3)=2\).
For point \(F\): \(x\)-coordinate of \(F\) is \(-5\), \(x\)-coordinate of \(F'\) is \(-3\). The change \(\Delta x=-3-(-5) = 2\).
For point \(G\): \(x\)-coordinate of \(G\) is \(1\), \(x\)-coordinate of \(G'\) is \(3\). The change \(\Delta x=3 - 1=2\).

Step2: Calculate the change in \(y\)-coordinate

For point \(E\): \(y\)-coordinate of \(E\) is \(4\), \(y\)-coordinate of \(E'\) is \(0\). The change \(\Delta y=0 - 4=-4\).
For point \(F\): \(y\)-coordinate of \(F\) is \(-1\), \(y\)-coordinate of \(F'\) is \(-5\). The change \(\Delta y=-5-(-1)=-4\).
For point \(G\): \(y\)-coordinate of \(G\) is \(1\), \(y\)-coordinate of \(G'\) is \(-3\). The change \(\Delta y=-3 - 1=-4\).

The translation rule \(T_{h,k}(x,y)=(x + h,y + k)\), where \(h = 2\) and \(k=-4\), so the rule is \(T_{2,-4}(x,y)\)

Answer:

\(T_{2,-4}(x,y)\)