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=-(1)/(2)x+b`

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

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

`\begin{gathered} -1=-(1)/(2)\cdot-3+b \\ -1=(3)/(2)+b \\ b=-1-(3)/(2)=-(5)/(2) \end{gathered}`

So, the equation of the required line:

In slope-intercept form is:

`y=-(1)/(2)x-(5)/(2)`

in standard form:

Multiply all terms by 2

`\begin{gathered} 2y=2\cdot-(1)/(2)x-2\cdot(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))=-(1)/(2)(x-(-3)) \\ \\ (y+1)=-(1)/(2)(x+3) \end{gathered}`