Question

Consider the line y = -7x+6.

Find the equation of the line that is parallel to this line and passes through the point (
5
5
)
Find the equation of the line that is perpendicular to this line and passes through the point (5,-5).

Answer 1

For this case we have that by definition, the equation of a line of the slope-intersection form is given by:

`y = mx + b`

Where:

m: It is the slope of the line

b: It is the cut-off point with the y axis

We have the following equation:

`y = -7x + 6`

With slope
`m_ {1} = - 7`

By definition, if two lines are parallel then their slopes are equal. Thus, a parallel line will be of the form:

`y = -7x + b`

We replace the point (5,5) through which the line passes and find "b":

`5 = -7 (5) + b\\5 = -35 + b\\5 + 35 = b\\b = 40`

Finally, the equation is:

`y = -7x + 40`

On the other hand, if two lines are perpendicular then the product of their slopes is -1. Thus, the slope of a perpendicular line will be:

`m_ {2} = \frac {-1} {m_ {1}} = \frac {-1} {- 7} = \frac {1} {7}`

Thus, the equation is of the form:

`y = \frac {1} {7} x + b`

We substitute the point (5, -5) through which the line passes and find "b":

`-5 = \frac {1} {7} (5) + b\\-5 = \frac {5} {7} + b\\-5- \frac {5} {7} = b\\b = \frac {-35-5} {7}\\b = \frac {-40} {7}\\b = - \frac {40} {7}\\`

Finally, the equation is:

`y = \frac {1} {7} x- \frac {40} {7}`

Answer:
`y = -7x + 40\\y = \frac {1} {7} x- \frac {40} {7}`