# Algebra: please check

posted by .

Find the equation of a line that passes through the point (1,2)
and has a slope of 1/3. Show how you found the equation.

Would the equation be:

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

Do I need to finish this? I am not sure how.

• Algebra: please check -

You set it up right. Don't forget your plugging in for x and y and you need to finish this.

So your equation would be

-2=1/3(1)+b
B would be your unknown y-intercept.
But this is not in slope-intercept form
So now you need to solve for b.

-2=1/3(1)+b
-2=1/3+b

subtract 1/3 from both sides you get
-2 1/3=b
or
-7/3=b

your equation would then be
y=1/3x-2 1/3
or
y=1/3x-7/3

• Algebra: please check -

There are several ways to write the equation of a straight line.
The way troy showed you is the y-intercept-slope form.
Your starting equation suggests that you have learned the point-slope form
continue from
y-2=1/3(x-1) by multiplying both sides by 3 to avoid fractions ...
3y - 6 = x-1
take everybody to the left side
-x + 3y - 5 = 0
usually we start with a positive x
x - 3y + 5 = 0

This is the general form of the equation of a straight line, and it contains no fractions.