Question

Write the equation of a line perpendicular to y=3x+1and goes through the point ( 6,2) y=−13x+4 y=−13x−4 y=3x+4 y=3x−4

Answer 1

`y=-(1)/(3)x+4`

Explanation:

step 1

Find the slope of the perpendicular line

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal

(the product of their slopes is equal to -1)

In this problem

we have

`y=3x+1`

The equation of the given line is
`m=3`

so

the slope of the perpendicular line to the given line is

`m=-(1)/(3)`

step 2

Find the equation of the line in point slope form

`y-y1=m(x-x1)`

we have

`m=-(1)/(3)`

`(x1,y1)=(6,2)`

substitute

`y-2=-(1)/(3)(x-6)`

Convert to slope intercept form

`y=mx+b`

Distribute right side

`y-2=-(1)/(3)x+2`

`y=-(1)/(3)x+2+2`

`y=-(1)/(3)x+4`