Question

Find the equation of the line that contains the point (-2, -1) and is perpendicular

Answer 1

Answer:

The line equation in the slope-intercept form:

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

Step-by-step explanation:

We know that the slope-intercept of line equation is

$y = mx+b$

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

Given the line

$2x + 3y = 9$

Writing in the slope-intercept form

$2x + 3y = 9$

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

Therefore, the slope of the line = m = -2/3

We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:

slope = m = -2/3

perpendicular slope = – 1/m

                                  $=-\frac{1}{-\frac{2}{3}}=\frac{3}{2}$

Given the point

(x₁, y₁) = (-2, -1)

Using the point-slope form of the line equation

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

where m is the slope and (x₁, y₁) is the point

substituting the perpendicular slope m = 3/2 and the point (-2, -1)

$y-\left(-1\right)=\frac{3}{2}\left(x-\left(-2\right)\right)$

Writing in the slope-intercept form

$y+1=\frac{3}{2}\left(x+2\right)$

subtract 1 from both sides

$y+1-1=\frac{3}{2}\left(x+2\right)-1$

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

Thus, the line equation in the slope-intercept form:

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