Question

question 5
1 pts
which set of coordinates provide the vertices for a 90° rotation of △abc about point a?
(image of a coordinate grid with triangle abc: a(-3,2), b(-1,2), c(-1,5))
options:
○ a(-3, 2), b(-1, 2), c(1, 1)
○ a(-3, -2), b(-1, -2), c(-1, 1)
○ a(-3, 2), b(-3, 0), c(0, 0)
○ a(-1, 4), b(-1, 2), c(2, 2)
question 6
1 pts
what are the coordinates of the image of p(3, -4) under a reflection in the x-axis?

Explanation:

Question 5 (Rotation of Triangle)

Step1: Find Original Coordinates

First, identify the original coordinates of \( A \), \( B \), and \( C \) from the graph.

• \( A(-3, 2) \) (since it's at \( x=-3 \), \( y=2 \))
• \( B(-1, 2) \) (at \( x=-1 \), \( y=2 \))
• \( C(-1, 5) \)? Wait, no—wait, looking at the graph, \( C \) is at \( (-1, 5) \)? Wait, no, the grid: \( A \) is at \( (-3, 2) \), \( B \) at \( (-1, 2) \), \( C \) at \( (-1, 5) \)? Wait, no, the y-axis: the grid lines. Wait, maybe I misread. Wait, the triangle: \( A \) is at \( (-3, 2) \), \( B \) at \( (-1, 2) \), \( C \) at \( (-1, 5) \)? No, the shaded triangle: \( A(-3,2) \), \( B(-1,2) \), \( C(-1,5) \)? Wait, no, the y-coordinate for \( C \) is 5? Wait, the graph has \( y \)-axis with +6, +5, etc. Wait, maybe the original \( C \) is \( (-1, 5) \), but we need a 90° rotation about \( A \).

Wait, 90° rotation about a point: the formula for rotating a point \( (x,y) \) 90° counterclockwise about \( (a,b) \) is \( (a - (y - b), b + (x - a)) \). Let's check the options.

First, let's list original points:

• \( A(-3, 2) \) (stays the same, since rotation is about \( A \))
• \( B(-1, 2) \): vector from \( A \) to \( B \) is \( (2, 0) \) (since \( -1 - (-3) = 2 \), \( 2 - 2 = 0 \)). Rotating this vector 90° counterclockwise: \( (0, 2) \) (because a 90° counterclockwise rotation of \( (x,y) \) is \( (-y, x) \); so \( (2,0) \) becomes \( (0, 2) \)). So new \( B \) is \( A + (0, 2) \)? Wait, no—wait, rotation about \( A \): the vector \( \overrightarrow{AB} \) is \( (2, 0) \). Rotating this vector 90° counterclockwise: the new vector \( \overrightarrow{AB'} \) should be \( (0, 2) \) (since 90° rotation of \( (a, b) \) is \( (-b, a) \); so \( (2,0) \) becomes \( (0, 2) \)). So \( B' = A + (0, 2) = (-3, 2 + 2) = (-3, 4) \)? No, that's not matching options. Wait, maybe clockwise? 90° clockwise rotation: vector \( (x,y) \) becomes \( (y, -x) \). So \( (2,0) \) becomes \( (0, -2) \). Then \( B' = A + (0, -2) = (-3, 2 - 2) = (-3, 0) \). Ah! Now check option 3: \( A(-3,2) \), \( B(-3,0) \), \( C(0,0) \). Let's check \( C \): original \( C \) is \( (-1, 5) \)? No, original \( C \) is \( (-1, 5) \)? Wait, no, original \( C \) is \( (-1, 5) \)? Wait, no, looking at the graph, \( C \) is at \( (-1, 5) \)? Wait, no, the y-coordinate for \( C \) is 5? Wait, the grid: each square is 1 unit. So \( A(-3,2) \), \( B(-1,2) \), \( C(-1,5) \). Vector \( \overrightarrow{AC} \) is \( (2, 3) \) (since \( -1 - (-3) = 2 \), \( 5 - 2 = 3 \)). Rotating this vector 90° clockwise: \( (3, -2) \) (since 90° clockwise rotation of \( (x,y) \) is \( (y, -x) \); so \( (2,3) \) becomes \( (3, -2) \)). Then new \( C \) is \( A + (3, -2) = (-3 + 3, 2 - 2) = (0, 0) \). Perfect! So \( B' \) is \( A + (0, -2) = (-3, 0) \), \( C' \) is \( (0, 0) \), and \( A \) stays \( (-3, 2) \). So the third option: \( A(-3,2) \), \( B(-3,0) \), \( C(0,0) \) matches.

Step2: Verify Option 3

Check the third option:

• \( A(-3, 2) \) (matches original \( A \), since rotation is about \( A \))
• \( B(-3, 0) \): original \( B(-1, 2) \), vector \( \overrightarrow{AB} = (2, 0) \). Rotating 90° clockwise: \( (0, -2) \), so \( A + (0, -2) = (-3, 2 - 2) = (-3, 0) \) (correct).
• \( C(0, 0) \): original \( C(-1, 5) \)? Wait, no—wait, original \( C \) is \( (-1, 5) \)? Wait, no, looking at the graph again: the triangle has \( A(-3,2) \), \( B(-1,2) \), \( C(-1,5) \)? No, the y-coordinate for \( C \) is 5? Wait, the grid lines: the vertical line at \( x=-1 \), and \( y=5 \) (since the dot is at \( y=5 \)). Then vector \( \overrightarrow{AC} = (-1 - (-3), 5 - 2) = (2, 3) \). Rot…

The rule for reflecting a point \( (x, y) \) over the \( x \)-axis is \( (x, -y) \). So for \( P(3, -4) \), the reflection will have the same \( x \)-coordinate and the \( y \)-coordinate negated.

Step1: Apply Reflection Rule

Given \( P(3, -4) \), reflect over \( x \)-axis: \( (3, -(-4)) = (3, 4) \).

Answer:

\( \boldsymbol{A(-3, 2), B(-3, 0), C(0, 0)} \) (the third option)

Question 6 (Reflection over x-axis)