Question

Consider the line y=6/7 x -7. Find the equation of the line that is perpendicular to this line and passes through the point (6,-6). Find the equation of the line that is parallel to this line and passes through the point (6,-6) .

Answer 1

`y= -\frac76x +1`;
`y= \frac67x -\frac{78}7`

Explanation:

We know from theory that two perpendicular lines have the product of their slope equal to -1. Let's find the second slope:

`\frac67m = -1 \rightarrow m=-\frac76`

At this point we can just use the slope-point equation to get our first line:

`y-y_0 = m(x-x_0)\\y-(-6) = -\frac76(x-6) \\y+6= -\frac76x +7\\y= -\frac76x +1`

Now, we know that two parallel lines have the same slope. Time to apply the same point-slope formula, with the original m this time.

`y-(-6) = \frac67(x-6)\\y+6 = \frac67x -\frac{36}7\\y= \frac67x -\frac{36}7 -{\frac{42}7`

`y= \frac67x -\frac{78}7`