Question

What is the equation, in slope-intercept form, of the line that is perpendicular to the line
y – 4 = –Two-thirds(x – 6) and passes through the point (−2, −2)?

y = –Two-thirdsx – Ten-thirds
y = –Two-thirdsx + Ten-thirds
y = Three-halvesx – 1
y = Three-halvesx + 1

Answer 1

The given line's equation is in point-slope form:

\[ y - 4 = -\frac{2}{3}(x - 6) \]

To find the slope of this line, we can see that the slope \( m \) is \( -\frac{2}{3} \).

Lines that are perpendicular to each other have slopes that are negative reciprocals. The negative reciprocal of \( -\frac{2}{3} \) is:

\[ m_{\text{perpendicular}} = \frac{3}{2} \]

Now, we want to find the equation of the line that has this slope and passes through the point \( (-2, -2) \). We will use the point-slope form of the equation of a line, which is:

\[ y - y_1 = m(x - x_1) \]

Substituting in the point \( (-2, -2) \) and the slope \( \frac{3}{2} \):

\[ y - (-2) = \frac{3}{2}(x - (-2)) \]
\[ y + 2 = \frac{3}{2}(x + 2) \]

Now, we can simplify and convert it to slope-intercept form \( y = mx + b \):

1. Distribute \( \frac{3}{2} \):

\[ y + 2 = \frac{3}{2}x + \frac{3}{2} \cdot 2 \]
\[ y + 2 = \frac{3}{2}x + 3 \]

2. Subtract \( 2 \) from both sides:

\[ y = \frac{3}{2}x + 3 - 2 \]
\[ y = \frac{3}{2}x + 1 \]

So, the equation of the line in slope-intercept form is:

\[ y = \frac{3}{2}x + 1 \]

Thus, the correct answer is:

**y = Three-halvesx + 1**