Question

Write the equation of the line that contains the point (−2, 7) and is

a. Parallel to x = 3.
b. Perpendicular to x = −3.
c. Parallel to y = 6x − 13.
d. Perpendicular to y = 6x − 13.

Answer 1

a.
`x=-2`

b.
`y=7`

c.
`y = 6x+c`

d.
`y=-(1)/(6)x+c`

Explanation:

part a: parallel to x = 3

`x = 3` it is a vertical line that passes through
`x = 3`, so a parallel line to this that passes through (-2,7) must be
`x=-2` which is also a vertical line (thus they are parallel because they will never touch).

part b: perpendicular to x = -3

`x = -3` it is a vertical line that passes through
`x = -3`, so a perpendicular line to this that passes through (-2,7) must be
`y = 7` because this is horizontal line, and since one line is vertical and the other is horizontal they are perpendicular.

part c: parallel to y = 6x - 13

For two lines to be parallel they must have the same slope. The slope of a line is the number that accompanies the x. for
`y=6x-13` the slope is 6. the equation for a parallel line will be:
`y = 6x+c` (where c can be any number) because the slope (6) is the same for both.

part d: perpendicular to y=6x-13

The condition for two lines to be perpendicular is:

`m_(1)m_(2)=-1`

where
`m_(1)` is the slope of one line, and
`m_(2)` the slope of the perpendicular line. We have the slope of the first line as we found in part c:
`m_(1)=6` so the slope of the second line has to be:

`m_(1)m_(2)=-1`

`6m_(2)=-1`

`m_(2)=-(1)/(6)`

so an equation of a line with a slope of
`-(1)/(6)` can be:

`y=-(1)/(6)x+c` (where c can be any number)