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