Question

What is the equation of a line that is perpendicular to y=-5x+11 and passes through the point (3,3)?

Answer 1

`y=(1)/(5)x+(12)/(5)`

Explanation:

The slope of a line that is perpendicular to another will have an opposite-reciprocal slope. So, if the slope was -2, then the perpendicular slope would be
`(1)/(2)` .

`y=-5x+11`

This is written in slope-intercept form:

`y=mx+b`

m is the slope and b is the y-intercept. Find the opposite-reciprocal of the slope, -5:

`y=(1)/(5) x+b`

Now we need to find the y-intercept. For this, substitute the given points for its appropriate value:

`(3_(x),3_(y))\\\\3=(1)/(5)(3)+b`

Solve for b:

Simplify multiplication:

`(1)/(5)*(3)/(1)=(3)/(5)`

Insert:

`3=(3)/(5)+b`

Subtract b from both sides:

`3-b=(3)/(5)+b-b\\\\3-b=(3)/(5)`

Subtract 3 from both sides:

`3-3-b=(3)/(5)-3\\\\-b=(3)/(5)-3`

Simplify subtraction:

`(3)/(5)-3\\\\(3)/(5)-(3)/(1)\\\\(3)/(5)-(15)/(5)=-(12)/(5)`

Insert:

`-b=-(12)/(5)`

Multiply both sides by - 1 to simplify b (it can be seen as -1b):

`-b*(-1)=-(12)/(5)*(-1)\\\\b=(12)/(5)`

Insert:

`y=(1)/(5)x+(12)/(5)`

:Done