Question

What is the equation, in slope-intercept form, of the line that is perpendicular to the line y – 4 = –2/3(x – 6) and passes through the point (−2, −2)?

Answer 1

y+2=3/2(x+2)
y+2=3/2x+3
y=3/2x+1

Answer 2

Answer:

The equation of required line is $y=\frac{3}{2}x+1$.

Step-by-step explanation:

If a line passes through a points $(x_1,y_1)$ with slope m, then the point slope form of the line is

$y-y_1=m(x-x_1)$

The given equation of line is

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

It means the slope of this line is $-\frac{2}{3}$.

Product of slopes of two perpendicular lines is -1. So, the slope of perpendicular line is $\frac{3}{2}$.

The slope of required line is $\frac{3}{2}$ and it passes through the point (-2,-2). So, the equation of line is

$y-(-2)=\frac{3}{2}(x-(-2))$

$y+2=\frac{3}{2}(x+2)$

$y+2=\frac{3}{2}x+3$

Subtract 2 from both sides.

$y=\frac{3}{2}x+3-2$

$y=\frac{3}{2}x+1$

Therefore, the equation of required line is $y=\frac{3}{2}x+1$.