Question

draw the image of quadrilateral (and triangle?) under a translation by 1 unit to the right and 0 units up.

Explanation:

Step1: Identify vertices of \( \triangle ABC \)

First, we need to find the coordinates of the vertices of triangle \( ABC \). Let's assume the coordinates of \( A \), \( B \), and \( C \) from the graph. From the graph, let's say:

• \( A=(1, - 4) \)
• \( B=(-7, - 2) \)
• \( C=(-5,1) \)

Step2: Apply the transformation (1 unit right, 6 units up)

The transformation rule for a point \( (x,y) \) is \( (x + 1,y + 6) \).

For point \( A=(1,-4) \):
New \( x=1 + 1=2 \), New \( y=-4 + 6 = 2 \). So \( A'=(2,2) \)

For point \( B=(-7,-2) \):
New \( x=-7 + 1=-6 \), New \( y=-2 + 6 = 4 \). So \( B'=(-6,4) \)

For point \( C=(-5,1) \):
New \( x=-5 + 1=-4 \), New \( y=1 + 6 = 7 \). So \( C'=(-4,7) \)

Step3: Plot the new points and draw the triangle

Plot the points \( A'(2,2) \), \( B'(-6,4) \), \( C'(-4,7) \) on the coordinate plane and connect them to get the image of \( \triangle ABC \) after the transformation.

(Note: Since this is a drawing - based problem, the key is to find the new coordinates of the vertices and then plot them. The actual drawing would involve marking these points on the grid and joining them as a triangle.)

Answer:

The image of \( \triangle ABC \) is obtained by transforming each vertex \( (x,y) \) to \( (x + 1,y + 6) \), getting \( A'(2,2) \), \( B'(-6,4) \), \( C'(-4,7) \), and then drawing the triangle with these vertices. (The actual drawing should be done on the given coordinate grid by plotting these new points and connecting them.)