Question

draw the image of \\( \triangle a b c \\) under a translation by 1 unit to the left and 5 units up.

Explanation:

Step1: Find the coordinates of the original triangle's vertices

Assume the coordinates of vertices \(A\), \(B\), \(C\) (from the graph). Let's say \(A(x_1,y_1)\), \(B(x_2,y_2)\), \(C(x_3,y_3)\). For a translation of \(1\) unit to the left (subtract \(1\) from \(x -\) coordinate) and \(5\) units up (add \(5\) to \(y -\) coordinate).
The translation rule is \((x,y)\to(x - 1,y + 5)\)

Step2: Calculate the new coordinates

For vertex \(A\): If \(A\) has coordinates \((x_1,y_1)\), then \(A'(x_1-1,y_1 + 5)\)
For vertex \(B\): If \(B\) has coordinates \((x_2,y_2)\), then \(B'(x_2-1,y_2 + 5)\)
For vertex \(C\): If \(C\) has coordinates \((x_3,y_3)\), then \(C'(x_3-1,y_3 + 5)\)

Step3: Plot the new vertices

Plot the points \(A'\), \(B'\), \(C'\) on the coordinate - plane and connect them to form the translated triangle \(\triangle A'B'C'\)

Answer:

Follow the steps above to draw the translated triangle \(\triangle A'B'C'\) with vertices obtained by applying the translation rule \((x,y)\to(x - 1,y + 5)\) to the vertices of \(\triangle ABC\)