Question

Use the table to answer the question.

Corrdinates of Line AB and CD
Line Point 1 Point 2
AB
(−3,6)
(3,8)
CD
(3,5)
(__,−1)
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 point)
Responses

5
5

−1
negative 1

−15
negative 15

1

Answer 1

To determine the missing coordinate for line CD such that the lines AB and CD are perpendicular, we first need to find the slopes of both lines.

### Step 1: Calculate the slope of line AB

The formula for the slope \(m\) between 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 AB with points \((-3, 6)\) and \((3, 8)\):

- \(x_1 = -3\), \(y_1 = 6\)
- \(x_2 = 3\), \(y_2 = 8\)

Now, plug in these values:

\[
m_{AB} = \frac{8 - 6}{3 - (-3)} = \frac{2}{3 + 3} = \frac{2}{6} = \frac{1}{3}
\]

### Step 2: Calculate the slope of line CD

Line CD has one point as \((3, 5)\) and the other point as \((__, -1)\). Let's denote the missing x-coordinate as \(x\).

Using the slope formula again for line CD:

\[
m_{CD} = \frac{-1 - 5}{x - 3} = \frac{-6}{x - 3}
\]

### Step 3: Determine the perpendicular condition

Two lines are perpendicular if the product of their slopes equals \(-1\):

\[
m_{AB} \cdot m_{CD} = -1
\]

Plugging in the slopes:

\[
\frac{1}{3} \cdot \frac{-6}{x - 3} = -1
\]

### Step 4: Solve for x

Rearranging the equation:

\[
\frac{-6}{3(x - 3)} = -1
\]

Multiplying both sides by \(3(x - 3)\):

\[
-6 = -3(x - 3)
\]

Expanding the right side:

\[
-6 = -3x + 9
\]

Adding \(3x\) to both sides:

\[
3x - 6 = 9
\]

Adding 6 to both sides:

\[
3x = 15
\]

Dividing by 3:

\[
x = 5
\]

### Conclusion

Therefore, the missing coordinate in the point for line CD is **5**.

The answer is 5.