Question

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

Answer 1

Answer:

The slope-intercept equation is:

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

Step-by-step explanation:

Given the equation

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

comparing it with the point-slope form of the line equation

$y-y_1=m\left(x-x_1\right)$

where m is the slope

• so the slope of the line is -2/3.

As we know that the slope of the perpendicular line is basically the negative reciprocal of the slope of the line, so

The slope of the perpendicular line will be: 3/2

The point-slope form of the equation of the perpendicular line that goes through (-2, -2) is:

$y-y_1=m\left(x-x_1\right)$

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

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

writing the line equation in the slope-intercept form

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

subtract 2 from both sides

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

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

Thus, the slope-intercept equation is:

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

Here,

As the slope-intercept form is

$y=mx+b$

where m is the slope and b is the y-intercept

so

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

m=3/2

b = y-intercept = 1

Therefore, the slope-intercept equation is:

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