Question

5. What is the equation of a line that is perpendicular to -x + 2y = 4 and passes through the point (-2,1)?
a.
Write the equation -x + 2y = 4 in slope-intercept form
b. Identify the slope of the line represented in part a.
C.
What is the slope of the line perpendicular to the line in steps a and b?
d. Write the equation of the perpendicular line in slope-intercept form.

Answer 1

Answer:

Please check the explanation.

Step-by-step explanation:

Given the equation

$-x + 2y = 4$

a) writing the equation in the slope-intercept form

We know that the slope-intercept form of the equation of the line is

$y=mx+b$

where m is the slope of the line

so writing the equation in the slope-intercept form

$-x + 2y = 4$

$2y=4+x$

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

b) Identify the slope of the line represented in part a

As the equation in slope-intercept form is

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

Here,

m = slope = 1/2     ∵ $y=mx+b$

c) What is the slope of the line perpendicular to the line in steps a and b

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

As the slope = 1/2

So the slope of the perpendicular line will be: -2

d. Write the equation of the perpendicular line in slope-intercept form.

Therefore, the point-slope form of the equation of the perpendicular line that goes through (-2,1) is:

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

substituting the values m = -2 and the point (-2,1)

$\:y-1=-2\left(x-\left(-2\right)\right)$

$y-1=-2\left(x+2\right)$

Add 1 to both sides

$y-1+1=-2\left(x+2\right)+1$

$y=-2x-3$