Question

On a coordinate plane, line A B goes through (negative 4, 0) and (2, negative 3). Point C is at (negative 2, 2).
Which point on the x-axis lies on the line that passes through point C and is parallel to line AB?

(1, 0)
(1, 1)
(0, 2)
(2, 0)

Answer 1

To find the point on the x-axis that lies on the line passing through point C (−2, 2) and is parallel to line AB, we first need to determine the slope of line AB.

Let's use the coordinates of points A (−4, 0) and B (2, −3) to calculate the slope (m) of line AB.

The formula for the slope between two points (x₁, y₁) and (x₂, y₂) is:
\[
m = \frac{y₂ - y₁}{x₂ - x₁}
\]

Substituting in the coordinates of A and B:
\[
m_{AB} = \frac{-3 - 0}{2 - (-4)} = \frac{-3}{2 + 4} = \frac{-3}{6} = -\frac{1}{2}
\]

So the slope of line AB is \( -\frac{1}{2} \).

Since the line we want to find through point C (−2, 2) is parallel to line AB, it will have the same slope of \( -\frac{1}{2} \).

Now, we can use the point-slope form of the line equation to find the equation of the line through point C:
\[
y - y₁ = m(x - x₁)
\]
Where \( (x₁, y₁) = (−2, 2) \) and \( m = -\frac{1}{2} \).

Substituting in these values:
\[
y - 2 = -\frac{1}{2}(x + 2)
\]

Expanding this:
\[
y - 2 = -\frac{1}{2}x - 1
\]
\[
y = -\frac{1}{2}x + 1
\]

Now we need to find the intersection of this line with the x-axis. On the x-axis, \( y = 0 \), so we set the equation equal to zero:
\[
0 = -\frac{1}{2}x + 1
\]

Solving for \( x \):
\[
\frac{1}{2}x = 1
\]
\[
x = 2
\]

Thus, the point on the x-axis that is parallel to line AB and passes through point C is \( (2, 0) \).

**Answer: (2, 0)**