Line AB = Point 1 - (4, 6) Point 2 - (3, 8)

Line CD = Point 1 - (1, 5) Point 2 - (___,4)
The table includes two points that fall on each of the lines, line AB and line CD. What must the value of the missing coordinate be in order to prove the lines are perpendicular?

1 answer

To determine the missing coordinate of point C in line CD such that line AB and line CD are perpendicular, we first need to find the slopes of both lines and set the product of their slopes equal to -1 (the condition for perpendicular lines).

Step 1: Calculate the slope of line AB

Let Point 1 be \( A(4, 6) \) and Point 2 be \( B(3, 8) \).

The formula for the slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

For line AB: \[ m_{AB} = \frac{8 - 6}{3 - 4} = \frac{2}{-1} = -2 \]

Step 2: Calculate the slope of line CD

Let Point 1 be \( C(1, 5) \) and Point 2 be \( D(x, 4) \) where \( x \) is the missing x-coordinate.

Using the slope formula again, we have: \[ m_{CD} = \frac{4 - 5}{x - 1} = \frac{-1}{x - 1} \]

Step 3: Set up the equation for the slopes to be perpendicular

The lines are perpendicular if: \[ m_{AB} \cdot m_{CD} = -1 \] Substituting in the slopes we found: \[ -2 \cdot \frac{-1}{x - 1} = -1 \]

Simplifying the equation: \[ \frac{2}{x - 1} = -1 \]

Step 4: Solve for \( x \)

Multiplying both sides by \( x - 1 \): \[ 2 = -1(x - 1) \] \[ 2 = -x + 1 \] Adding \( x \) to both sides: \[ x + 2 = 1 \] Subtracting 2 from both sides: \[ x = 1 - 2 = -1 \]

Thus, the missing coordinate for point D in line CD is: \[ \boxed{-1} \]