Write an equation for a line that is perpendicular to an equation and passes through a point?

Question

Hi, i am sorry i am working on a summer packet and i totally forgot how to do some of these things

anyhelp would be really appreciated
I dont need the answer, just a quick explanation on how to do it

the question is: Write an equation for the line that is perpendicular to 9x+3y=6 and that passes through the point (-1,5)
I tried to put 9x+3y=6 into mx+b form but coudlnt do that either
i really need to go back to school
THANKS FOR YOUR HELP

Answer 1

Putting 9x + 3y = 6 into slope-intercept form:
y = -3x + 2.
That has gradient -3, and lines perpendicular to it have gradient 1/3.

You then find the equation of a line with gradient 1/3 which passes through (-1,5):
(y - 5) / (x + 1) = 1/3
Simplify that, and you are home.

Answer 2

Given an equation of a line 9 x + 3y = 6
find an equation for a line that is perpendicular to an equation and passes through a point (-1, 5)

the slope of the given line is - 9/3 = -3
so the slope of the perpendicular line is 1/3
y = a + (1/3) x
when x = -1 y =5
5 = a - 1/3
a = 5 & 1/3 = 16/3

y = 16/3 + (1/3)x

Answer 3

a) divide all the equation by 3 : 3x +y = 6
b) now isolate y: y = -3x + 6 ... the angular coefficient of this line is m =-3
c) the line you want must have an angular coefficient m1 so that: m1(m) = -1 So, m1 must be 1/3
d) the line you want is y = m1x + b and m1 = 1/3 ,
so y = x/3 + b
e) To find b, just put the point into the equation:
5 = -1/3 + b ==> b = 5+1/3 = 16/3
f) Finally the equation is y = x/3 + 16/3 or multiplying the equation by 3 x-3y + 16 = 0

Answer 4

You are off to the right start. To do so, solve the equation for 1*y. You get y= 2-3x. This equation has a slope of -3. The perpendicular has a slope of +(1/3). Sooooo your mystery equation is
y = 1/3 x + b. Substitute x=-1 and y=5 in this equation and solve for b. Ta,da.