Question

randy draws triangle abc on the coordinate plane with vertices a(7,-4), b(10,3), and c(6,1). he then translates the figure so the coordinates of the image are a(5,1), b(8,8), and c(4,6). what rule did he use to draw the image?
t_{-5,2}(x,y)
t_{-2,5}(x,y)
t_{2,-5}(x,y)
t_{5,-2}(x,y)

Explanation:

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

For point \(A\): \(x\) - coordinate of \(A\) is \(7\), \(x\) - coordinate of \(A'\) is \(5\). The change \(\Delta x=5 - 7=-2\).
For point \(B\): \(x\) - coordinate of \(B\) is \(10\), \(x\) - coordinate of \(B'\) is \(8\). The change \(\Delta x = 8-10=-2\).
For point \(C\): \(x\) - coordinate of \(C\) is \(6\), \(x\) - coordinate of \(C'\) is \(4\). The change \(\Delta x=4 - 6=-2\).

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

For point \(A\): \(y\) - coordinate of \(A\) is \(-4\), \(y\) - coordinate of \(A'\) is \(1\). The change \(\Delta y=1-(-4)=5\).
For point \(B\): \(y\) - coordinate of \(B\) is \(3\), \(y\) - coordinate of \(B'\) is \(8\). The change \(\Delta y=8 - 3=5\).
For point \(C\): \(y\) - coordinate of \(C\) is \(1\), \(y\) - coordinate of \(C'\) is \(6\). The change \(\Delta y=6 - 1=5\).

The translation rule \(T_{a,b}(x,y)=(x + a,y + b)\), where \(a=-2\) and \(b = 5\). So the rule is \(T_{-2,5}(x,y)\).

Answer:

B. \(T_{- 2,5}(x,y)\)