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



Answer :

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

1. Identify the slope of the given line:
The given line equation is in point-slope form:
[tex]\[ y - 4 = -\frac{2}{3}(x - 6) \][/tex]
The coefficient of [tex]\((x - 6)\)[/tex] is the slope of the line. So, the slope [tex]\( m_1 \)[/tex] of the given line is:
[tex]\[ m_1 = -\frac{2}{3} \][/tex]

2. Find the slope of the perpendicular line:
The slope of a line perpendicular to another is the negative reciprocal of the original slope. Therefore, the slope [tex]\( m_2 \)[/tex] of the perpendicular line is:
[tex]\[ m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} \][/tex]

3. Use the slope-intercept form (y = mx + b) to find the y-intercept:
The point [tex]\((-2, -2)\)[/tex] lies on the perpendicular line. Using this point and the slope [tex]\( \frac{3}{2} \)[/tex], we can find the y-intercept [tex]\( b \)[/tex]:

[tex]\[ y = mx + b \][/tex]
Substituting [tex]\((x_1, y_1) = (-2, -2)\)[/tex] and [tex]\( m_2 = \frac{3}{2} \)[/tex]:

[tex]\[ -2 = \frac{3}{2}(-2) + b \][/tex]

4. Solve for the y-intercept [tex]\( b \)[/tex]:
Calculate [tex]\( \frac{3}{2}(-2) \)[/tex]:

[tex]\[ \frac{3}{2} \times -2 = -3 \][/tex]

Substitute back into the equation to solve for [tex]\( b \)[/tex]:

[tex]\[ -2 = -3 + b \][/tex]

Adding 3 to both sides:

[tex]\[ b = 1 \][/tex]

5. Write the final equation:
With [tex]\( m_2 = \frac{3}{2} \)[/tex] and [tex]\( b = 1 \)[/tex], the equation of the line in slope-intercept form is:

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

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

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