Question

Write a linear equation of a line that passes (-3,-1) and is perpendicular to the graph of y = 2x+3 in eitherslope-intercept, point-slope, or standard form.

Answer 1

The general slope intercept form of the line is

y = mx + b

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

So,

For the given equation y = 2x + 3

the slope = m = 2

Now, we need to find the equation of the line which is perpendicular to the given line and pass through the point (-3 , -1 )

The slope of the required line = m' = -1/2

Because the product of the two slopes = -1

so, the equation of the required line will be:

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

find the value of b using the point (-3 , -1 )

so, when x = -3 , y = -1

$\begin{gathered} -1=-\frac{1}{2}\cdot-3+b \\ -1=\frac{3}{2}+b \\ b=-1-\frac{3}{2}=-\frac{5}{2} \end{gathered}$

So, the equation of the required line:

In slope-intercept form is:

$y=-\frac{1}{2}x-\frac{5}{2}$

in standard form:

Multiply all terms by 2

$\begin{gathered} 2y=2\cdot-\frac{1}{2}x-2\cdot\frac{5}{2} \\ 2y=-x-5 \\ \\ x+2y=-5 \end{gathered}$

Finally, in point - slope form

The slope is -1/2 and the point is ( -3 , -1 )

So, the equation will be:

$\begin{gathered} (y-(-1))=-\frac{1}{2}(x-(-3)) \\ \\ (y+1)=-\frac{1}{2}(x+3) \end{gathered}$