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

Answer:

$y=-\frac{1}{3}x+4$

Step-by-step 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

$m=-\frac{1}{3}$

step 2

Find the equation of the line in point slope form

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

we have

$m=-\frac{1}{3}$

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

substitute

$y-2=-\frac{1}{3}(x-6)$

Convert to slope intercept form

$y=mx+b$

Distribute right side

$y-2=-\frac{1}{3}x+2$

$y=-\frac{1}{3}x+2+2$

$y=-\frac{1}{3}x+4$