Question

draw the line of reflection that reflects $\triangle abc$ onto $\triangle abc$.

Explanation:

Step1: Identify corresponding points

Find the midpoint between each pair of corresponding points (e.g., \( B(4,4) \) and \( B'(-4,4) \), \( A(6,-6) \) and \( A'(-4,-6) \), \( C(0,-2) \) and \( C'(2,-2) \)). Wait, actually, for \( B \) and \( B' \): \( B \) is at \( (4,4) \), \( B' \) at \( (-4,4) \). The midpoint's x - coordinate is \( \frac{4 + (-4)}{2}=0 \), y - coordinate \( \frac{4 + 4}{2}=4 \)? Wait no, wait \( B \) and \( B' \): looking at the grid, \( B \) is at \( (4,4) \), \( B' \) at \( (-4,4) \). The midpoint between \( (x_1,y_1) \) and \( (x_2,y_2) \) is \( (\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2}) \). So for \( B(4,4) \) and \( B'(-4,4) \), midpoint is \( (\frac{4-4}{2},\frac{4 + 4}{2})=(0,4) \)? Wait no, \( 4+(-4)=0 \), divided by 2 is 0, y - coordinate 4. For \( C(0,-2) \) and \( C'(2,-2) \), midpoint is \( (\frac{0 + 2}{2},\frac{-2-2}{2})=(1,-2) \)? Wait, maybe better to check the vertical or horizontal line. Wait, \( B \) and \( B' \) have the same y - coordinate (4), so the line of reflection should be the vertical line that is the perpendicular bisector of \( BB' \). Since \( BB' \) is horizontal (same y), the perpendicular bisector is vertical. The midpoint of \( BB' \): \( B \) is at \( x = 4 \), \( B' \) at \( x=-4 \), so midpoint x - coordinate is \( \frac{4+(-4)}{2}=0 \), y - coordinate 4. So the vertical line \( x = 0 \)? Wait no, wait \( B \) is at (let's re - check the grid: the blue dots for \( B \) and \( B' \): \( B \) is at (4,4), \( B' \) at (-4,4). So the distance between them is 8 units (from x=-4 to x = 4). The midpoint is at x = 0 (since - 4 and 4 are 8 units apart, midpoint at 0). So the line of reflection is the vertical line \( x = 0 \)? Wait, but let's check another pair. \( A \) and \( A' \): \( A \) is at (6,-6), \( A' \) at (-4,-6). Midpoint of \( A(6,-6) \) and \( A'(-4,-6) \): x - coordinate \( \frac{6-4}{2}=1 \), y - coordinate \( \frac{-6-6}{2}=-6 \). Wait, that's different. Wait, maybe I misread the coordinates. Wait the red dot \( A' \) is at (-4,-6), blue dot \( A \) at (6,-6). So midpoint x: (6 + (-4))/2 = 1, y: (-6 + (-6))/2=-6. \( C \) is at (0,-2), \( C' \) at (2,-2). Midpoint x: (0 + 2)/2 = 1, y: (-2 + (-2))/2=-2. Oh! So \( A \) and \( A' \) midpoint x = 1, y=-6; \( C \) and \( C' \) midpoint x = 1, y=-2. So the line of reflection is the vertical line \( x = 1 \)? Wait, let's recalculate:

For \( A(6,-6) \) and \( A'(-4,-6) \):

Midpoint \( x=\frac{6+(-4)}{2}=\frac{2}{2}=1 \), \( y=\frac{-6+(-6)}{2}=-6 \)

For \( C(0,-2) \) and \( C'(2,-2) \):

Midpoint \( x=\frac{0 + 2}{2}=1 \), \( y=\frac{-2+(-2)}{2}=-2 \)

For \( B(4,4) \) and \( B'(-4,4) \):

Midpoint \( x=\frac{4+(-4)}{2}=0 \), \( y = 4 \). Wait, this is a contradiction. Wait, maybe I misidentified \( B \) and \( B' \). Wait the blue dots: \( B \) is on the right, \( B' \) on the left. Let's look at the labels: \( B \) is next to the blue dot on the right, \( B' \) on the left. Wait, maybe \( B \) is at (4,4), \( B' \) at (-4,4). But \( A \) is at (6,-6), \( A' \) at (-4,-6). Wait, the distance between \( A \) and \( A' \): 6 - (-4)=10 units? No, 6 to - 4 is 10 units? Wait 6 - (-4)=10, midpoint at (6-4)/2 = 1. So x = 1. For \( C \): \( C \) is at (0,-2), \( C' \) at (2,-2). Midpoint at (1,-2). So the line \( x = 1 \) is the vertical line that is the perpendicular bisector of \( AA' \), \( CC' \). Let's check \( BB' \): if \( B \) is at (4,4) and \( B' \) at (-2,4)? Wait, maybe my initial coordinate reading was wrong. Let's re - examine the grid:

- The x - axis and y - axis: the vertical lines are x=-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7. The horizontal lines are y=-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7.

- \( B \) (blue dot) is at (4,4), \( B' \) (blue dot) is at (-4,4)? No, wait the distance between them: from x=-4 to x = 4 is 8 units. But \( A \) is at (6,-6), \( A' \) at (-4,-6): distance is 10 units, midpoint at x = 1. \( C \) is at (0,-2), \( C' \) at (2,-2): midpoint at x = 1. So the line of reflection is \( x = 1 \).

Wait, let's calculate the midpoint for \( BB' \) again. If \( B \) is at (4,4) and \( B' \) is at (-2,4) (maybe I misread - 4 as - 2). Let's check the grid: the blue dot for \( B' \) is at x=-4? No, looking at the grid, the left blue dot is at x=-4, y = 4; the right blue dot is at x = 4, y = 4. Then \( A \) is at x = 6, y=-6; \( A' \) at x=-4, y=-6. So the midpoint of \( AA' \): (6 + (-4))/2=1, (-6 + (-6))/2=-6. Midpoint of \( CC' \): (0 + 2)/2=1, (-2 + (-2))/2=-2. Midpoint of \( BB' \): (4 + (-4))/2=0, (4 + 4)/2=4. Wait, this is inconsistent. There must be a mistake in my coordinate reading.

Wait, the key property of a reflection: the line of reflection is the perpendicular bisector of the segment joining a point and its image. So for each pair of corresponding points (A and A', B and B', C and C'), the line of reflection is the perpendicular bisector of \( AA' \), \( BB' \), and \( CC' \).

Let's take \( A(6,-6) \) and \( A'(-4,-6) \):

- The slope of \( AA' \): since y - coordinates are the same (-6), slope is 0 (horizontal line). So the perpendicular bisector is vertical (undefined slope). The midpoint of \( AA' \) is \( (\frac{6+(-4)}{2},\frac{-6+(-6)}{2})=(1,-6) \). So the perpendicular bisector is the vertical line \( x = 1 \).

Take \( C(0,-2) \) and \( C'(2,-2) \):

- Slope of \( CC' \) is 0 (horizontal line), so perpendicular bisector is vertical. Midpoint of \( CC' \) is \( (\frac{0 + 2}{2},\frac{-2+(-2)}{2})=(1,-2) \). So the vertical line \( x = 1 \).

Take \( B(4,4) \) and \( B'(-2,4) \) (maybe I misread \( B' \) as - 4, but looking at the grid, \( B' \) is at x=-2? Wait, the blue dot for \( B' \) is at x=-4? No, the grid lines: from the left, the blue dot is at x=-4, y = 4; right blue dot at x = 4, y = 4. But \( A \) is at x = 6, y=-6; \( A' \) at x=-4, y=-6. Then midpoint of \( AA' \) is (1,-6). Midpoint of \( BB' \) is (0,4). That can't be. Wait, maybe the triangles are labeled differently. Wait the problem says "reflects \( \triangle ABC \) onto \( \triangle A'B'C' \)". So \( A \) maps to \( A' \), \( B \) to \( B' \), \( C \) to \( C' \).

Looking at the coordinates:

- \( A \): (6,-6)

- \( A' \): (-4,-6)

- \( B \): (4,4)

- \( B' \): (-2,4)

- \( C \): (0,-2)

- \( C' \): (2,-2)

Now recalculate midpoints:

- Midpoint of \( AA' \): \( (\frac{6+(-4)}{2},\frac{-6+(-6)}{2})=(1,-6) \)

- Midpoint of \( BB' \): \( (\frac{4+(-2)}{2},\frac{4 + 4}{2})=(1,4) \)

- Midpoint of \( CC' \): \( (\frac{0+2}{2},\frac{-2+(-2)}{2})=(1,-2) \)

Ah! Now all midpoints have x - coordinate 1. So the line of reflection is the vertical line \( x = 1 \), since all midpoints lie on \( x = 1 \), and the segments \( AA' \), \( BB' \), \( CC' \) are horizontal (slope 0), so their perpendicular bisectors are vertical (slope undefined) passing through x = 1.

Step2: Draw the line

The line of reflection is the vertical line \( x = 1 \). To draw it, we can plot the line \( x = 1 \) (a vertical line passing through (1,y) for all y - values).

Answer:

The line of reflection is the vertical line \( x = 1 \) (or the line passing through \( x = 1 \) on the coordinate grid). To draw it, sketch a vertical line at \( x = 1 \) that passes through the midpoints of the segments joining corresponding vertices (\( AA' \), \( BB' \), \( CC' \)).