Question

Which equation describes a line that passes through (-6,8) and is perpendicular to the line described by y = 2x - 42A. y= -2x - 4B. y= -1/2x + 5C. y= 1/2 + 11D. y= 2x + 20

Answer 1

we must find a line perpendicular to

`y=2x-42`

which passes through (-6,8). As we can see, this given lines has slope m=2.

First, the slope of a perpendicular line is the reciprocal inverse of the given line. In other words,

the perpendicular lines must have slope M equal to

`M=-(1)/(m)`

In our case m=2, hence, the perpendicular line has slope

`M=-(1)/(2)`

Therefore, the perpendicular line has the form

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

and now, we must find the y-intercept b. This can be done by substituying the given point (-6,8)

into the last equation:

`8=-(1)/(2)(-6)+b`

which gives

`\begin{gathered} 8=(6)/(2)+b \\ 8=3+b \\ b=8-3 \\ b=5 \end{gathered}`

Finally, the answer is

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