Question

the figure below is dilated by a factor of 3 centered at the origin. plot the resulting image.
click twice to plot a segment.
click a segment to delete it.

Explanation:

Step1: Identify Coordinates

First, find the coordinates of each vertex of the original figure. Let's assume the coordinates (from the graph):
- Left vertical segment: Let's say the top point is \( (-9, 9) \) and bottom is \( (-9, 0) \) (since it's on x=-9, y from 0 to 9).
- For the quadrilateral (R, Q, P, O, N):
- \( R(-3, 0) \) (on x=-3, y=0)
- \( Q(-3, 2) \) (x=-3, y=2)
- \( P(3, 0) \) (x=3, y=0)
- \( O(3, -1) \) (x=3, y=-1)
- \( N(0, -2) \)? Wait, maybe better to check grid. Wait, the grid has x from -10 to 10, y from -10 to 10. Let's re-express:
- Left vertical line: top at (-9, 9), bottom at (-9, 0) (since it's a vertical segment from ( -9, 0) to (-9, 9)).
- For the other figure: R is at (-3, 0), Q at (-3, 2), P at (3, 0), O at (3, -2)? Wait, maybe N is at (0, -2)? Wait, maybe I misread. Let's correct:
Let's list each vertex:
1. Left vertical segment: points A(-9, 0) and B(-9, 9) (assuming the black dot is at (-9,9) and bottom at (-9,0)).
2. Quadrilateral: R(-3, 0), Q(-3, 2), P(3, 0), O(3, -2), N(0, -2)? Wait, maybe the coordinates are:
- R: (-3, 0)
- Q: (-3, 2)
- P: (3, 0)
- O: (3, -2)
- N: (0, -2)
Wait, maybe better to use the grid. Each square is 1 unit. So:
- Left vertical line: from (-9, 0) to (-9, 9) (since it's 9 units up from x=-9, y=0 to y=9).
- For the other shape (the quadrilateral with R, Q, P, O, N):
- R: (-3, 0) (x=-3, y=0)
- Q: (-3, 2) (x=-3, y=2)
- P: (3, 0) (x=3, y=0)
- O: (3, -2) (x=3, y=-2)
- N: (0, -2) (x=0, y=-2)
- Wait, maybe N is (0, -2) and O is (3, -2), P is (3, 0), Q is (-3, 2), R is (-3, 0).

Step2: Apply Dilation

Dilation centered at origin with factor 3: multiply each coordinate by 3.
- For left vertical segment:
- A(-9, 0) → A'(-9*3, 0*3) = (-27, 0)
- B(-9, 9) → B'(-9*3, 9*3) = (-27, 27)
- For quadrilateral vertices:
- R(-3, 0) → R'(-3*3, 0*3) = (-9, 0)
- Q(-3, 2) → Q'(-3*3, 2*3) = (-9, 6)
- P(3, 0) → P'(3*3, 0*3) = (9, 0)
- O(3, -2) → O'(3*3, -2*3) = (9, -6)
- N(0, -2) → N'(0*3, -2*3) = (0, -6)

Wait, maybe I made a mistake in original coordinates. Let's re-express the original figure:

Looking at the graph:

- The left vertical segment: the top point is at x=-9, y=9 (since it's 9 units above x-axis, x=-9). The bottom is at x=-9, y=0 (on x-axis).

- The other figure (the polygon with R, Q, P, O, N):

- R is at x=-3, y=0 (on x-axis, 3 units left of origin).

- Q is at x=-3, y=2 (2 units up from R).

- P is at x=3, y=0 (3 units right of origin, on x-axis).

- O is at x=3, y=-2 (2 units down from P).

- N is at x=0, y=-2 (on the y-axis, 2 units down, connected to O).

So dilation by 3: multiply each x and y by 3.

So:

- Left vertical segment:

- Top: (-9, 9) → (-27, 27)

- Bottom: (-9, 0) → (-27, 0)

- Quadrilateral:

- R(-3, 0) → (-9, 0)

- Q(-3, 2) → (-9, 6)

- P(3, 0) → (9, 0)

- O(3, -2) → (9, -6)

- N(0, -2) → (0, -6)

Now, to plot the image:

- Draw the vertical segment from (-27, 0) to (-27, 27).

- Draw the quadrilateral with vertices (-9, 0), (-9, 6), (9, 0), (9, -6), (0, -6), and back to (-9, 0)? Wait, no, the original quadrilateral is R-Q-P-O-N-R, so the sides are RQ, QP, PO, ON, NR? Wait, RQ is vertical, QP is a line from Q(-3,2) to P(3,0), PO is vertical down, ON is horizontal left, NR is vertical up? Wait, maybe the original figure is a pentagon? Wait, maybe the original figure has vertices R(-3,0), Q(-3,2), P(3,0), O(3,-2), N(0,-2), and back to R? Wait, no, connecting R to Q to P to O to N to R? Let's check the slopes:

Q(-3,2) to P(3,0): slope is (0-2)/(3 - (-3)) = (-2)/6 = -1/3.

P(3,0) to O(3,-2): vertical line (x=3, y from 0 to -2).

O(3,-2) to N(0,-2): horizontal line (y=-2, x from 3 to 0).

N(0,-2) to R(-3,0): slope is (0 - (-2))/(-3 - 0) = 2/(-3) = -2/3. Wait, maybe my initial coordinate for N is wrong. Maybe N is at (0, -1)? Wait, the grid has y from -10 to 10, each square is 1 unit. Let's count the grid lines.

Alternatively, maybe the original coordinates are:

- Left vertical segment: ( -9, 0 ) to ( -9, 9 )

- Quadrilateral:

- R: ( -3, 0 )

- Q: ( -3, 2 )

- P: ( 3, 0 )

- O: ( 3, -2 )

- N: ( 0, -2 )

So dilation by 3:

- Left segment: ( -27, 0 ) to ( -27, 27 )

- Quadrilateral vertices:

- R': ( -9, 0 )

- Q': ( -9, 6 )

- P': ( 9, 0 )

- O': ( 9, -6 )

- N': ( 0, -6 )

Now, to plot these points:

1. For the left vertical segment: plot (-27, 0) and (-27, 27), then draw a vertical line between them.

2. For the quadrilateral: plot (-9, 0), (-9, 6), (9, 0), (9, -6), (0, -6), and connect them in order (R' to Q' to P' to O' to N' to R').

Wait, but maybe the original figure's left vertical segment is actually part of the same figure? No, the left vertical segment is a separate vertical line, and the other is a polygon.

Alternatively, maybe the original figure has two parts: a vertical line from (-9,0) to (-9,9), and a polygon with vertices (-3,0), (-3,2), (3,0), (3,-2), (0,-2).

After dilation by 3, the vertical line becomes from (-27,0) to (-27,27), and the polygon becomes (-9,0), (-9,6), (9,0), (9,-6), (0,-6).

To plot the image, you would:

- For the vertical line: mark the points (-27, 0) and (-27, 27) and draw a vertical segment between them.

- For the polygon: mark the points (-9, 0), (-9, 6), (9, 0), (9, -6), (0, -6) and connect them in the same order as the original (R to Q to P to O to N to R).

So the resulting image will have:

- A vertical segment at x = -27, from y=0 to y=27.

- A polygon with vertices at (-9, 0), (-9, 6), (9, 0), (9, -6), (0, -6).

Answer:

To plot the dilated image:

1. Left Vertical Segment: Plot points \((-27, 0)\) and \((-27, 27)\), then draw a vertical segment between them.
2. Quadrilateral (RQPON): Plot the dilated vertices:

• \( R'(-9, 0) \), \( Q'(-9, 6) \), \( P'(9, 0) \), \( O'(9, -6) \), \( N'(0, -6) \).

Connect these points in the order \( R' \to Q' \to P' \to O' \to N' \to R' \) to form the dilated polygon.

(Note: Coordinates are derived by multiplying original coordinates by 3, centered at the origin.)