Question

On a coordinate plane, line A B goes through (negative 2, 4) and (2, negative 8). Point C is at (6, 4).
Which point on the y-axis lies on the line that passes through point C and is perpendicular to line AB?

(–6, 0)
(0, –6)
(0, 2)
(2, 0)

Answer 1

To find the point on the y-axis that lies on the line through point C and is perpendicular to line AB, we need to follow these steps:

1. **Find the slope of line AB**: The slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
For points A(-2, 4) and B(2, -8):
\[
m_{AB} = \frac{-8 - 4}{2 - (-2)} = \frac{-12}{4} = -3
\]

2. **Determine the slope of the perpendicular line**: The slope of a line that is perpendicular to another line is the negative reciprocal of the original line's slope. Hence, the slope of the line through point C that is perpendicular to line AB is:
\[
m_{\perpendicular} = -\frac{1}{m_{AB}} = -\frac{1}{-3} = \frac{1}{3}
\]

3. **Find the equation of the line through point C (6, 4)**: We can use the point-slope form of the equation of a line, which is:
\[
y - y_1 = m(x - x_1)
\]
Plugging in point C (6, 4) and the perpendicular slope \(\frac{1}{3}\):
\[
y - 4 = \frac{1}{3}(x - 6)
\]
Simplifying this gives:
\[
y - 4 = \frac{1}{3}x - 2 \quad \Rightarrow \quad y = \frac{1}{3}x + 2
\]

4. **Find the intersection of this line with the y-axis**: The y-axis is where \( x = 0 \). Substituting \( x = 0 \) into the equation:
\[
y = \frac{1}{3}(0) + 2 = 2
\]
Hence, the point on the y-axis is \( (0, 2) \).

5. **Verify the options**: Among the given options, the point that lies on the y-axis is \( (0, 2) \).

Thus, the point on the y-axis that lies on the line passing through point C and perpendicular to line AB is:

\[
\boxed{(0, 2)}
\]