Question

On a coordinate plane, lines G H and G prime H prime are shown. Line G H has points (negative 2, 4) and (4, negative 5). Line G prime H prime has points (4, 2) and (negative 5, negative 4).
Are lines GH and G'H' 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.

Answer 1

To determine whether lines GH and G'H' are perpendicular, we need to find the slopes of both lines and check if they are negative reciprocals of each other.

**Step 1: Calculate the slope of line GH**

The slope \( m \) of a line through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

For line GH, using the points \((-2, 4)\) and \((4, -5)\):

\[
m_{GH} = \frac{-5 - 4}{4 - (-2)} = \frac{-9}{4 + 2} = \frac{-9}{6} = -\frac{3}{2}
\]

**Step 2: Calculate the slope of line G'H'**

For line G'H', using the points \((4, 2)\) and \((-5, -4)\):

\[
m_{G'H'} = \frac{-4 - 2}{-5 - 4} = \frac{-6}{-9} = \frac{2}{3}
\]

**Step 3: Check if the slopes are negative reciprocals**

Two lines are perpendicular if the slopes \( m_1 \) and \( m_2 \) satisfy:

\[
m_1 \cdot m_2 = -1
\]

Now, we check:

\[
m_{GH} \cdot m_{G'H'} = \left(-\frac{3}{2}\right) \cdot \left(\frac{2}{3}\right) = -\frac{6}{6} = -1
\]

Since \( m_{GH} \cdot m_{G'H'} = -1 \), the slopes are indeed negative reciprocals.

**Conclusion**

Yes, their slopes are negative reciprocals, so the answer is:

**Yes, their slopes are negative reciprocals.**