Question

What is an equation of the line that passes through the point (-4, -6) and is

perpendicular to the line 2x -y = 6?

Answer 1

The required equation is:
`\mathbf{x+2y=-16}`

Explanation:

We need to find equation of the line that passes through the point (-4, -6) and is perpendicular to the line

The equation of line will be in the slope intercept form
`y=mx+b`

where m is slope and b is y-intercept.

Finding Slope

First transforming the given equation
`2x -y = 6` in slope-intercept form:

`2x -y = 6\\-y=-2x+6\\y=2x-6`

Now comparing the equation with general equation of slope intercept form We get slope m= 2

Since the two lines are perpendicular, the slope of new line will be opposite reciprocal of given line i.e
`m=-(1)/(m)`

So, Slope of required line is:
`m=-(1)/(2)`

Finding y-intercept

y-intercept can be found using slope
`m=-(1)/(2)` and point (-4,-6) as follows

`y=mx+b\\-6=-(1)/(2)(-4)+b\\-6=+2+b\\b=-6-2\\b=-8`

So, y-intercept b = -8

Finding equation of line:

The equation of line having slope
`m=-(1)/(2)` and y-intercept b = -8 is:

`y=mx+b\\y=-(1)/(2)(x)-8`

Now, writing the equation in standard form i.e Ax+By=C

`y=-(1)/(2)(x)-8 \\ Multiply \ by \ 2\\2y=-x-16\\x+2y=-16`

So, the required equation is:
`\mathbf{x+2y=-16}`