What is the equation, in slope-intercept form, of the line that is perpendicular to the line [tex]\( y - 4 = -\frac{2}{3}(x - 6) \)[/tex] and passes through the point [tex]\((-2, -2)\)[/tex]?

A. [tex]\( y = -\frac{2}{3}x - \frac{10}{3} \)[/tex]
B. [tex]\( y = -\frac{2}{3}x + \frac{10}{3} \)[/tex]
C. [tex]\( y = \frac{3}{2}x - 1 \)[/tex]
D. [tex]\( y = \frac{3}{2}x + 1 \)[/tex]



Answer :

To find the equation of the line that is perpendicular to the line [tex]\( y - 4 = -\frac{2}{3}(x - 6) \)[/tex] and passes through the point [tex]\((-2, -2)\)[/tex], we need to follow these steps:

1. Determine the slope of the given line:

The equation is in point-slope form [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( m \)[/tex] is the slope. For the given line [tex]\( y - 4 = -\frac{2}{3}(x - 6) \)[/tex], the slope [tex]\( m \)[/tex] is [tex]\(-\frac{2}{3}\)[/tex].

2. Find the slope of the perpendicular line:

The slope of a line perpendicular to another is the negative reciprocal of the original line's slope. Therefore, the perpendicular slope [tex]\( m_{\perp} \)[/tex] will be:
[tex]\[ m_{\perp} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} \][/tex]

3. Use the point-slope form to determine the equation of the new line:

We need the equation of the line that passes through the point [tex]\((-2, -2)\)[/tex] with slope [tex]\( \frac{3}{2} \)[/tex]. The point-slope form is:
[tex]\[ y - y_1 = m_{\perp}(x - x_1) \][/tex]
Plugging in the point [tex]\((-2, -2)\)[/tex] and the slope [tex]\( \frac{3}{2} \)[/tex], we get:
[tex]\[ y - (-2) = \frac{3}{2}(x - (-2)) \][/tex]
Simplify this:
[tex]\[ y + 2 = \frac{3}{2}(x + 2) \][/tex]

4. Convert to slope-intercept form [tex]\( y = mx + b \)[/tex]:

Distribute [tex]\( \frac{3}{2} \)[/tex] on the right-hand side:
[tex]\[ y + 2 = \frac{3}{2}x + \frac{3}{2} \cdot 2 \][/tex]
[tex]\[ y + 2 = \frac{3}{2}x + 3 \][/tex]

Subtract 2 from both sides to isolate [tex]\( y \)[/tex]:
[tex]\[ y = \frac{3}{2}x + 3 - 2 \][/tex]
[tex]\[ y = \frac{3}{2}x + 1 \][/tex]

So, the equation of the line in slope-intercept form that is perpendicular to the given line and passes through the point [tex]\((-2, -2)\)[/tex] is:

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

The correct choice is:
[tex]\[ \boxed{y = \frac{3}{2} x + 1} \][/tex]