Consider the line y=5x-3. Find the equation of the line that is parallel to this line and passes through the point (2, -5).
Find the equation of the line that is perpendicular to this line and passes through the point (2, -5).

y=5x-15, y=-1/5x-23/5

Step-by-step explanation:

Parallel lines have the same slope, so we know the line that is parallel to y=5x-3 must have slope 5.

We now have y=5x+b. We are given the point (2, -5) so substitute the x and y coordinates into the equation.

y=5x+b

-5=5(2)+b

-5=10+b

b=-15

So the equation of the parallel line is y=5x-15.

Perpendicular lines must have slopes that multiply to -1, so we know the line perpendicular to y=5x-3 must have slope -1/5.

We now have y=-1/5x+b. We are given the point (2, -5) so substitute the x and y coordinates into the equation.

y=-1/5x+b

-5=-1/5(2)+b

-5=-2/5+b

b=-23/5

So the equation of the perpendicular line is y=-1/5x-23/5.

the answer is b.)

< 1=< 2

its 50 miles away from the starting point