Question

graph the image of \triangle abc after a translation 3 units up.

Explanation:

Step1: Find the coordinates of original points

Assume the coordinates of \(A\), \(B\), \(C\) are \(A(x_1,y_1)\), \(B(x_2,y_2)\), \(C(x_3,y_3)\). From the graph (assuming \(A(3, - 7)\), \(B(8,-7)\), \(C(-1,-10)\))

Step2: Apply the translation rule

The rule for a translation \(3\) units up is \((x,y)\to(x,y + 3)\).
For point \(A\): \((3,-7)\to(3,-7 + 3)=(3,-4)\)
For point \(B\): \((8,-7)\to(8,-7 + 3)=(8,-4)\)
For point \(C\): \((-1,-10)\to(-1,-10 + 3)=(-1,-7)\)

Step3: Plot the new points and connect them

Plot \(A'(3,-4)\), \(B'(8,-4)\), \(C'(-1,-7)\) on the coordinate - plane and connect them to form \(\triangle A'B'C'\)

Answer:

Plot the points \((3,-4)\), \((8,-4)\), \((-1,-7)\) and connect them to get the image of \(\triangle ABC\) after a translation \(3\) units up.