Question

are lines gh and gh perpendicular? why or why not?
○ yes, they intersect at one point.
○ yes, their slopes are negative reciprocals.
○ no, their slopes are not negative reciprocals.
○ no, their slopes are not equal.

Explanation:

Step1: Find coordinates of points

For line \( GH \): \( G(-2, 4) \), \( H(4, -5) \)
For line \( G'H' \): \( G'(4, 2) \), \( H'(-4, -4) \)

Step2: Calculate slope of \( GH \)

Slope formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \)
\( m_{GH} = \frac{-5 - 4}{4 - (-2)} = \frac{-9}{6} = -\frac{3}{2} \)

Step3: Calculate slope of \( G'H' \)

\( m_{G'H'} = \frac{-4 - 2}{-4 - 4} = \frac{-6}{-8} = \frac{3}{4} \)? Wait, no, recheck \( H' \) coordinates. Wait, \( H' \) is at \( (-4, -4) \)? Wait, looking at the graph, \( H' \) is at \( (-5, -4) \)? Wait, maybe I misread. Wait, the blue line: \( G'(4, 2) \), \( H'(-5, -4) \)? Wait, no, let's re-express. Wait, the grid: \( G \) is at \( (-2, 4) \), \( H \) at \( (4, -5) \). \( G' \) at \( (4, 2) \), \( H' \) at \( (-5, -4) \)? Wait, no, the blue line: from \( H'(-5, -4) \) to \( G'(4, 2) \). So slope \( m_{G'H'} = \frac{2 - (-4)}{4 - (-5)} = \frac{6}{9} = \frac{2}{3} \). Wait, earlier mistake. Let's recalculate \( GH \): \( G(-2, 4) \), \( H(4, -5) \): \( \frac{-5 - 4}{4 - (-2)} = \frac{-9}{6} = -\frac{3}{2} \). \( G'H' \): \( G'(4, 2) \), \( H'(-5, -4) \): \( \frac{2 - (-4)}{4 - (-5)} = \frac{6}{9} = \frac{2}{3} \). Wait, no, maybe \( H \) is at \( (4, -5) \), \( G \) at \( (-2, 4) \). Slope of \( GH \): \( \frac{4 - (-5)}{-2 - 4} = \frac{9}{-6} = -\frac{3}{2} \). Slope of \( G'H' \): \( G'(4, 2) \), \( H'(-4, -4) \)? Wait, the graph: \( H' \) is at \( (-5, -4) \)? Maybe I misread. Wait, the key is: two lines are perpendicular if their slopes are negative reciprocals (product is -1). Let's check the correct slopes. Wait, maybe the correct coordinates: \( G(-2, 4) \), \( H(4, -5) \): slope \( m_1 = \frac{-5 - 4}{4 - (-2)} = -\frac{9}{6} = -\frac{3}{2} \). \( G'(4, 2) \), \( H'(-5, -4) \): slope \( m_2 = \frac{2 - (-4)}{4 - (-5)} = \frac{6}{9} = \frac{2}{3} \). Wait, \( -\frac{3}{2} \times \frac{2}{3} = -1 \). Oh! So they are negative reciprocals. Wait, maybe my initial coordinate for \( H' \) was wrong. Let's re-express: \( G' \) is at \( (4, 2) \), \( H' \) is at \( (-5, -4) \)? No, looking at the grid, \( H' \) is at \( (-5, -4) \)? Wait, the blue line: from \( H'(-5, -4) \) to \( G'(4, 2) \). So the slope is \( (2 - (-4))/(4 - (-5)) = 6/9 = 2/3 \). The orange line: \( G(-2, 4) \) to \( H(4, -5) \): slope is \( (-5 - 4)/(4 - (-2)) = -9/6 = -3/2 \). Now, \( (-3/2) \times (2/3) = -1 \), so they are negative reciprocals. Therefore, the lines are perpendicular because their slopes are negative reciprocals.

Answer:

Yes, their slopes are negative reciprocals. (Corresponding option: "Yes, their slopes are negative reciprocals.")