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.
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
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.
Yes, you are on the right track with the equation:
y - 2 = (1/3)(x - 1)
To finish getting the equation, you can simplify and rearrange the equation to put it in the standard form, which is y = mx + b, where m is the slope and b is the y-intercept.
Let's simplify the equation:
y - 2 = (1/3)(x - 1)
First, distribute (1/3) to (x - 1):
y - 2 = (1/3)x - 1/3
Next, let's isolate y by adding 2 to both sides:
y - 2 + 2 = (1/3)x - 1/3 + 2
Simplifying further:
y = (1/3)x + 5/3
So, the equation of the line that passes through the point (1, 2) with a slope of 1/3 is y = (1/3)x + 5/3.
Now you have the complete equation of the line.