Question

draw the image of quadrilateral abcd under a translation by 1 unit to the right and 4 units up.

Explanation:

Step1: Identify original coordinates

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

• \(A\): Let's assume \(A\) is at \((-2, -5)\) (wait, looking at the graph, actually, let's re - check. Wait, the blue points: Wait, the original quadrilateral \(ABCD\): Let's find the coordinates properly. Let's see, point \(B\) is at \((-4, 3)\), point \(A\) is at \((-2, -5)\)? Wait, no, maybe I misread. Wait, the graph: Let's look at the coordinates. Let's list the vertices:
• \(A\): From the graph, the bottom - left vertex (the one labeled \(A\)): Let's check the x - coordinate and y - coordinate. The x - coordinate is \(-2\), y - coordinate is \(-5\)? Wait, no, maybe the original coordinates: Let's see, the translation is 1 unit right and 4 units up. Let's correctly identify the original coordinates:
• Let's assume the original vertices:
• \(A\): Let's say \(A(-2,-5)\) (wait, no, looking at the blue dots, maybe the original quadrilateral has vertices: Let's re - examine. Wait, the problem says quadrilateral \(3CD\) (probably a typo, should be \(ABCD\)). Let's find the coordinates of \(A\), \(B\), \(C\), \(D\):
• \(B\): \((-4, 3)\)
• \(A\): Let's see, the point at \((-2,-5)\) (the bottom - most left vertex)
• \(C\): \((6, 2)\) (wait, no, the right - top vertex \(C\) is at \((6, 2)\)? Wait, no, the y - coordinate for \(C\) is 2? Wait, no, looking at the vertical line for \(C\), it's at \(x = 6\), and the horizontal line: the y - coordinate for \(C\) is 2? Wait, no, the blue line for \(C\) is at \(y = 2\)? Wait, no, maybe the original coordinates are:
• \(A(-2,-5)\), \(B(-4,3)\), \(C(6,2)\), \(D(6,-2)\) (wait, the bottom - right vertex \(D\) is at \((6,-2)\))

Step2: Apply translation rule

The translation rule for a point \((x,y)\) translated \(h\) units right and \(k\) units up is \((x + h,y + k)\). Here, \(h = 1\) (right) and \(k = 4\) (up).

• For point \(A(x_{A},y_{A})\):
• If \(A\) is \((-2,-5)\), then the translated \(A'\) is \((-2 + 1,-5+4)=(-1,-1)\)
• For point \(B(x_{B},y_{B})=(-4,3)\):
• Translated \(B'\) is \((-4 + 1,3 + 4)=(-3,7)\)
• For point \(C(x_{C},y_{C})=(-6,2)\)? Wait, I think I made a mistake in identifying \(C\)'s coordinate. Let's re - identify the original coordinates correctly. Let's look at the grid:
• Let's take the correct original coordinates:
• \(B\): \((-4,3)\) (x=-4, y = 3)
• \(A\): Let's say \(A(-2,-5)\) (x=-2, y=-5)
• \(C\): \((6,2)\) (x = 6, y = 2)
• \(D\): \((6,-2)\) (x = 6, y=-2)
• Now apply translation \((x + 1,y + 4)\):
• For \(A(-2,-5)\): \(A'=(-2 + 1,-5 + 4)=(-1,-1)\)
• For \(B(-4,3)\): \(B'=(-4+1,3 + 4)=(-3,7)\)
• For \(C(6,2)\): \(C'=(6 + 1,2+4)=(7,6)\)
• For \(D(6,-2)\): \(D'=(6 + 1,-2 + 4)=(7,2)\)

Step3: Plot the translated points

After finding the translated coordinates of \(A'\), \(B'\), \(C'\), \(D'\), we plot these points on the coordinate plane and connect them to get the image of the quadrilateral after translation.

(Note: Since this is a graph - based problem, the main steps are identifying the original coordinates, applying the translation formula \((x,y)\to(x + 1,y + 4)\) to each vertex, and then plotting the new points. The key is to correctly identify the original coordinates of each vertex of the quadrilateral and then apply the translation rule.)

Answer:

To draw the image of quadrilateral \(ABCD\) under the translation \(1\) unit right and \(4\) units up:

1. Find the original coordinates of \(A\), \(B\), \(C\), \(D\) (e.g., \(A(-2,-5)\), \(B(-4,3)\), \(C(6,2)\), \(D(6,-2)\) from the graph).
2. Apply the translation \((x,y)\to(x + 1,y + 4)\) to each vertex:

• \(A'(-2 + 1,-5 + 4)=(-1,-1)\)
• \(B'(-4+1,3 + 4)=(-3,7)\)
• \(C'(6 + 1,2+4)=(7,6)\)
• \(D'(6 + 1,-2 + 4)=(7,2)\)

1. Plot the points \(A'\), \(B'\), \(C'\), \(D'\) and connect them to form the translated quadrilateral.

(If we were to describe the final graph, the translated quadrilateral will have vertices at \((-1,-1)\), \((-3,7)\), \((7,6)\), and \((7,2)\), and the sides will be parallel to the original quadrilateral's sides as translation is a rigid transformation.)