Question

On a coordinate plane, a line goes through (0, negative 1) and (3, 1). A point is at (negative 3, 0).
What is the equation of the line that is parallel to the given line and has an x-intercept of –3?

y = Two-thirdsx + 3
y = Two-thirdsx + 2
y = Negative three-halvesx + 3
y = –Three-halvesx + 2

Answer 1

To find the equation of a line that is parallel to a given line and has an x-intercept of -3, we first need to determine the slope of the original line that runs through the points (0, -1) and (3, 1).

1. **Calculate the slope of the original line:**

The slope (m) is calculated using the formula:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
where (x_1, y_1) = (0, -1) and (x_2, y_2) = (3, 1).

\[
m = \frac{1 - (-1)}{3 - 0} = \frac{1 + 1}{3} = \frac{2}{3}
\]

So, the slope of the original line is \( \frac{2}{3} \).

2. **Since we want a parallel line, it will have the same slope.**

Thus, the slope of the line we are looking for is also \( \frac{2}{3} \).

3. **Find the equation of the parallel line with an x-intercept of -3:**

An x-intercept of -3 means that when \( x = -3 \), \( y = 0 \). We can use the point-slope form of the equation of a line:
\[
y - y_1 = m(x - x_1)
\]
where \( (x_1, y_1) \) is the point (-3, 0) and \( m = \frac{2}{3} \).

Substituting in the values:
\[
y - 0 = \frac{2}{3}(x - (-3))
\]
\[
y = \frac{2}{3}(x + 3)
\]

4. **Simplify the equation:**
\[
y = \frac{2}{3}x + \frac{2}{3} \cdot 3
\]
\[
y = \frac{2}{3}x + 2
\]

So, the equation of the parallel line with an x-intercept of -3 is:
\[
y = \frac{2}{3}x + 2
\]

Thus, the correct answer is:
**y = Two-thirdsx + 2**.