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write the coordinates of the vertices after a reflection over the line …

Question

write the coordinates of the vertices after a reflection over the line ( y = x ).

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of points \( F \), \( G \), and \( H \) from the graph.

  • For point \( F \): It lies on the \( y \)-axis at \( (0, 6) \) (since \( x = 0 \) and \( y = 6 \)).
  • For point \( G \): It lies on the \( y \)-axis at \( (0, 9) \) (wait, looking at the graph, the \( y \)-coordinate for \( G \) is 9? Wait, the grid: the \( y \)-axis has 6 at \( F \), and \( G \) is above, let's check again. Wait, the graph: \( F \) is at \( (0, 6) \), \( G \) is at \( (0, 9) \)? Wait, no, the grid lines: each square is 1 unit. Let's re - examine:
  • Point \( F \): \( x = 0 \), \( y = 6 \), so \( F(0,6) \)
  • Point \( G \): \( x = 0 \), \( y = 9 \)? Wait, the graph shows \( G \) at \( (0,9) \)? Wait, no, the vertical axis: the top of \( G \) is at \( y = 9 \)? Wait, maybe I misread. Wait, the problem's graph: \( F \) is at \( (0,6) \), \( G \) is at \( (0,9) \)? Wait, no, looking at the blue line: \( F \) is at \( (0,6) \), \( G \) is at \( (0,9) \)? Wait, no, the \( y \)-axis: the first mark after 6 is 8, then 10. Wait, \( G \) is at \( (0,9) \)? Wait, maybe the original coordinates:
  • \( F \): \( (0,6) \)
  • \( G \): \( (0,9) \)
  • \( H \): \( (- 10,7) \) (since it's at \( x=-10 \), \( y = 7 \))

Step2: Apply reflection over \( y = x \)

The rule for reflecting a point \( (x,y) \) over the line \( y=x \) is to swap the \( x \) and \( y \) coordinates, i.e., the image of \( (x,y) \) is \( (y,x) \).

  • For point \( F(0,6) \):

After reflection over \( y = x \), we swap \( x \) and \( y \). So the new coordinates \( F' \) will be \( (6,0) \).

  • For point \( G(0,9) \):

Applying the reflection rule \( (x,y)\to(y,x) \), the new coordinates \( G' \) will be \( (9,0) \).

  • For point \( H(-10,7) \):

Applying the reflection rule, we swap \( x \) and \( y \). So the new coordinates \( H' \) will be \( (7,-10) \).

Answer:

\( F(6, 0) \)
\( G(9, 0) \)
\( H(7, - 10) \)

Wait, wait, I think I made a mistake in the original \( G \) coordinate. Let's re - check the graph. The \( y \)-axis: \( F \) is at \( (0,6) \), \( G \) is at \( (0,9) \)? Wait, no, the grid: the vertical line (y - axis) has \( F \) at \( (0,6) \), and \( G \) is at \( (0,9) \)? Wait, maybe the original \( G \) is at \( (0,9) \)? Wait, the problem's graph: the blue line from \( H(-10,7) \) to \( G(0,9) \) and \( H(-10,7) \) to \( F(0,6) \). Let's confirm the original coordinates:

  • \( H \): \( x=-10 \), \( y = 7 \), so \( H(-10,7) \)
  • \( F \): \( x = 0 \), \( y = 6 \), so \( F(0,6) \)
  • \( G \): \( x = 0 \), \( y = 9 \), so \( G(0,9) \)

Then, reflection over \( y=x \):

  • For \( F(0,6) \), reflection is \( (6,0) \)
  • For \( G(0,9) \), reflection is \( (9,0) \)
  • For \( H(-10,7) \), reflection is \( (7,-10) \)