Question

Line m contains the points -3,4 and 1,0. write the equation of a line that would be perpendicular to this one and pass through the point -2,6

How do you solve that

Answer 1

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

`y = mx + b`

Where:

m: It's the slope

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

According to the statement we have two points through which the line passes:

`(x_ {1}, y_ {1}): (- 3,4)\\(x_ {2}, y_ {2}) :( 1,0)`

We found the slope:

`m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {0-4} {1 - (- 3)} = \frac {-4} { 1 + 3} = \frac {-4} {4} = - 1`

By definition, if two lines are perpendicular then the product of their slopes is -1.

Thus, a perpendicular line will have a slope:
`m_ {2} = \frac {-1} {- 1} = 1`

Thus, the equation will be of the form:

`y = x + b`

We substitute the given point and find "b":

`6 = -2 + b\\6 + 2 = b\\b = 8`

Finally, the equation is:

`y = x + 8`

Answer:

`y = x + 8`