Question

What is the equation in slope-intercept form of the line that passes through the point (6, 3) and is parallel to the graph of y=−23x+12
?

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Answer 1

To find the equation of the line that passes through the point (6, 3) and is parallel to the line represented by the equation \( y = -23x + 12 \), we start by noting that parallel lines have the same slope.

1. **Identify the slope of the given line**:
The slope (m) of the line \( y = -23x + 12 \) is \(-23\).

2. **Use the point-slope form for the point (6, 3)**:
We can use the point-slope form of the equation of a line, which is given by:
\[
y - y_1 = m(x - x_1)
\]
where \((x_1, y_1)\) is the point on the line and \(m\) is the slope. Here, we have:
\((x_1, y_1) = (6, 3)\) and \(m = -23\).

3. **Substitute the values into the point-slope equation**:
\[
y - 3 = -23(x - 6)
\]

4. **Distribute the slope**:
\[
y - 3 = -23x + 138
\]

5. **Solve for y to put the equation into slope-intercept form**:
\[
y = -23x + 138 + 3
\]
\[
y = -23x + 141
\]

Thus, the equation of the line in slope-intercept form is:
\[
\boxed{y = -23x + 141}
\]