Question

What is an equation of the line that passes through the point (1,6) and is

perpendicular to the line x + 3y = 272

Answer 1

y = 3x + 3

Explanation:

Rules of Perpendicularity

`y = mx + b` ⊥
`y=-(1)/(m) x + b`

1 - Find your first equation

x + 3y = 272 | 1.1 - get y by itself by first subtracting x

-x -x

3y = -x + 272 | 1.2 - divide both sides by 3 and you've got it simplified

`y = -(1)/(3)x + (272)/(3)`

That 272 over 3 looks pretty intimidating, however, all we really care about is that slope

2 - Use the slope to find the perpendicular

`y = -(1)/(3)x+b` ⊥
`y= -\frac{1}{-\frac {1}{3}}x+ b = (3*1)/(1) x+b = 3x+b`

`y = -(1)/(3)x +b` ⊥
`y = 3x + b`

3 - Use the perpendicular y = mx + b to find b when it passes through said point

y = 3x + b must pass through (1, 6) or (x, y)

Plug x and y in

6 = 3(1) + b

6 = 3 + b

-3 -3

3 = b

4.) Rewrite y=mx+b form with the found b

y = 3x + 3

Hope this helps,

- J.M.Jamon

Pennsylvania State University

Full-time B.S.CE. undergrad student