Question

use the drawing tool(s) to form the correct answer on the provided graph.
a dilation by a scale factor of 2 centered at (2, -1) is performed on the triangle shown.
draw the resulting triangle.
(the image part is omitted as per instruction, only the text part is retained)

Explanation:

Step1: Identify Original Triangle Vertices

First, find the coordinates of the original triangle's vertices from the graph. Let's assume the vertices are \( A(-3, -1) \), \( B(2, -1) \), and \( C(1, 2) \) (wait, no, looking at the graph: the left vertex is at \( (-3, -1) \)? Wait, no, the y-coordinate for the base: the bottom vertices are at \( (-3, -1) \)? Wait, the graph shows: one vertex at \( (-3, -1) \) (wait, no, the y-axis: the bottom line is y=-1? Wait, the original triangle: let's check the coordinates. The left vertex: x=-3, y=-1? Wait, no, the point at (-3, -1)? Wait, the bottom two vertices: one at (-3, -1) and one at (2, -1)? Wait, no, the right bottom vertex is at (2, -1)? Wait, the top vertex is at (1, 2)? Wait, maybe better to list the three vertices:

Looking at the graph:

- Left vertex: \( (-3, -1) \) (wait, no, the y-coordinate: the grid lines. Let's see: the bottom horizontal line is y=-1? Wait, the original triangle has three vertices: let's call them \( P(-3, -1) \), \( Q(2, -1) \), and \( R(1, 2) \). Wait, no, the right bottom vertex is at (2, -1)? Wait, the x-axis: from -10 to 10, y from -10 to 10. The bottom two vertices: one at (-3, -1) (x=-3, y=-1) and one at (2, -1) (x=2, y=-1), and the top vertex at (1, 2) (x=1, y=2). Wait, no, the top vertex is at (1, 2)? Wait, the graph shows a triangle with vertices at (-3, -1), (2, -1), and (1, 2). Wait, maybe I misread. Let's check again:

Wait, the left vertex: x=-3, y=-1 (since it's on the line y=-1, x=-3). The right bottom vertex: x=2, y=-1 (on y=-1, x=2). The top vertex: x=1, y=2 (since it's above, x=1, y=2). So original vertices: \( A(-3, -1) \), \( B(2, -1) \), \( C(1, 2) \).

Step2: Apply Dilation Formula

The dilation formula centered at \( (h, k) = (2, -1) \) with scale factor \( r = 2 \) is:

For a point \( (x, y) \), the new point \( (x', y') \) is given by:

\( x' = h + r(x - h) \)

\( y' = k + r(y - k) \)

So the center is \( (2, -1) \), scale factor 2.

Step3: Dilate Each Vertex

Dilate Vertex \( A(-3, -1) \):

\( x' = 2 + 2(-3 - 2) = 2 + 2(-5) = 2 - 10 = -8 \)

\( y' = -1 + 2(-1 - (-1)) = -1 + 2(0) = -1 \)

So \( A'(-8, -1) \)

Dilate Vertex \( B(2, -1) \):

\( x' = 2 + 2(2 - 2) = 2 + 0 = 2 \)

\( y' = -1 + 2(-1 - (-1)) = -1 + 0 = -1 \)

So \( B'(2, -1) \) (wait, that's the same? No, wait, \( B \) is at (2, -1), so \( x - h = 2 - 2 = 0 \), so \( x' = 2 + 2(0) = 2 \), \( y' = -1 + 2(0) = -1 \). So \( B \) is on the center, so it remains the same? Wait, no, the center is (2, -1), so if a point is at the center, dilation doesn't move it. Wait, but the original triangle: is \( B \) at (2, -1)? Let's check the graph again. The right bottom vertex is at (2, -1), which is the center (2, -1)? Wait, the center of dilation is (2, -1), so that vertex is the center, so it stays.

Dilate Vertex \( R(1, 2) \):

\( x' = 2 + 2(1 - 2) = 2 + 2(-1) = 2 - 2 = 0 \)

\( y' = -1 + 2(2 - (-1)) = -1 + 2(3) = -1 + 6 = 5 \)

Wait, that can't be right. Wait, maybe I misidentified the original vertices. Let's re-express the original vertices correctly.

Wait, looking at the graph:

- Left vertex: \( (-3, -1) \) (x=-3, y=-1)

- Right bottom vertex: \( (2, -1) \) (x=2, y=-1)

- Top vertex: \( (1, 2) \) (x=1, y=2)

Wait, no, the top vertex is at (1, 2)? Wait, the x-coordinate: between -1 and 2? Wait, maybe the top vertex is at (1, 2). Let's confirm the center of dilation is (2, -1). So the vector from center (2, -1) to each vertex:

For vertex \( A(-3, -1) \): vector is \( (-3 - 2, -1 - (-1)) = (-5, 0) \). Multiply by scale factor 2: new vector is \( (-10, 0) \). So new vertex \( A' = (2 + (-10), -1 + 0) = (-8, -1) \).

For vertex \( B(2, -1) \): vector is \( (2 - 2, -1 - (-1)) = (0, 0) \). Multiply by 2: (0, 0). So new vertex \( B' = (2 + 0, -1 + 0) = (2, -1) \) (same as original, since it's the center).

For vertex \( C(1, 2) \): vector is \( (1 - 2, 2 - (-1)) = (-1, 3) \). Multiply by 2: (-2, 6). So new vertex \( C' = (2 + (-2), -1 + 6) = (0, 5) \).

Wait, but that seems off. Maybe the original top vertex is at (1, 2)? Wait, maybe I made a mistake in the original vertices. Let's check again. The original triangle: the bottom two vertices are at (-3, -1) and (2, -1), and the top vertex at (1, 2). So after dilation:

- \( A(-3, -1) \) becomes \( A' = (2 + 2*(-3 - 2), -1 + 2*(-1 - (-1))) = (2 - 10, -1 + 0) = (-8, -1) \)

- \( B(2, -1) \) becomes \( B' = (2, -1) \) (center)

- \( C(1, 2) \) becomes \( C' = (2 + 2*(1 - 2), -1 + 2*(2 - (-1))) = (2 - 2, -1 + 6) = (0, 5) \)

Wait, but maybe the original vertices are different. Let's re-express the original triangle's vertices correctly. Let's look at the graph again:

The left vertex is at (-3, -1) (x=-3, y=-1), the right bottom vertex at (2, -1) (x=2, y=-1), and the top vertex at (1, 2) (x=1, y=2). So the sides: from (-3, -1) to (2, -1) (length 5 units), from (2, -1) to (1, 2) (slope (2 - (-1))/(1 - 2) = 3/(-1) = -3), and from (1, 2) to (-3, -1) (slope (-1 - 2)/(-3 - 1) = (-3)/(-4) = 3/4).

After dilation with scale factor 2 centered at (2, -1):

- The side from (-3, -1) to (2, -1) (length 5) will be doubled, so from (-8, -1) to (2, -1) (length 10).

- The side from (2, -1) to (1, 2) (vector (-1, 3)) will be doubled to (-2, 6), so the new vertex is (2 - 2, -1 + 6) = (0, 5).

- The side from (1, 2) to (-3, -1) (vector (-4, -3)) will be doubled to (-8, -6), but wait, no, dilation is from the center, so we use the center to each vertex.

Wait, maybe a better approach: for each vertex \( (x, y) \), the new coordinates are \( (h + r(x - h), k + r(y - k)) \), where \( (h, k) = (2, -1) \), \( r = 2 \).

So let's list the original vertices correctly:

Looking at the graph:

- Vertex 1: (-3, -1) (left bottom)

- Vertex 2: (2, -1) (right bottom, center)

- Vertex 3: (1, 2) (top)

Now apply dilation:

Vertex 1: \( x' = 2 + 2*(-3 - 2) = 2 - 10 = -8 \), \( y' = -1 + 2*(-1 - (-1)) = -1 + 0 = -1 \) → (-8, -1)

Vertex 2: \( x' = 2 + 2*(2 - 2) = 2 + 0 = 2 \), \( y' = -1 + 2*(-1 - (-1)) = -1 + 0 = -1 \) → (2, -1) (same as original, since it's the center)

Vertex 3: \( x' = 2 + 2*(1 - 2) = 2 - 2 = 0 \), \( y' = -1 + 2*(2 - (-1)) = -1 + 6 = 5 \) → (0, 5)

Now, the resulting triangle will have vertices at (-8, -1), (2, -1), and (0, 5). To draw this, plot these three points and connect them with line segments.

Step2: Plot New Vertices and Draw Triangle

Using the drawing tools, plot the points (-8, -1), (2, -1), and (0, 5), then connect them with line segments to form the dilated triangle.

Answer:

The resulting triangle has vertices at \((-8, -1)\), \((2, -1)\), and \((0, 5)\). Draw these points and connect them. (Note: Since this is a drawing task, the key is to apply the dilation formula to each vertex and plot the new points.)