Question

draw the image of quadrilateral abcd under a translation by 2 units to the right and 5 units down.

Explanation:

Step1: Identify coordinates of ABCD

First, find the coordinates of each vertex of quadrilateral \(ABCD\). From the graph:

• \(A\): Let's assume \(A\) is at \((-4, 1)\) (by looking at the grid, x=-4, y=1)
• \(B\): \(B\) is at \((3, -1)\) (x=3, y=-1)
• \(C\): \(C\) is at \((4, 1)\) (x=4, y=1)
• \(D\): \(D\) is at \((-5, 3)\) (x=-5, y=3)

Step2: Apply translation rule

A translation of 2 units to the right (add 2 to the x - coordinate) and 5 units down (subtract 5 from the y - coordinate) has the rule \((x,y)\to(x + 2,y-5)\)

For point \(A(-4,1)\):

New \(x=-4 + 2=-2\), New \(y = 1-5=-4\). So \(A'(-2,-4)\)

For point \(B(3,-1)\):

New \(x=3 + 2 = 5\), New \(y=-1-5=-6\). So \(B'(5,-6)\)

For point \(C(4,1)\):

New \(x=4 + 2=6\), New \(y=1 - 5=-4\). So \(C'(6,-4)\)

For point \(D(-5,3)\):

New \(x=-5+2=-3\), New \(y=3 - 5=-2\). So \(D'(-3,-2)\)

Step3: Draw the image

Plot the points \(A'(-2,-4)\), \(B'(5,-6)\), \(C'(6,-4)\), \(D'(-3,-2)\) and connect them in order to get the translated quadrilateral \(A'B'C'D'\)

Answer:

The translated quadrilateral has vertices \(A'(-2,-4)\), \(B'(5,-6)\), \(C'(6,-4)\), \(D'(-3,-2)\) (and the image is drawn by plotting these points and connecting them).