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Equation of line in standard form
- Thread starter Mike
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- #2
- Jan 26, 2012
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Hi Mike,
Welcome to MHB!
There is something called the "point-slope" form of a line. It looks like this. \(\displaystyle y-y_0=m(x-x_0)\). $m$ is the slope and the $x_0$ and $y_0$ refer to a point $(x_0,y_0)$.
If we plug in your info we get \(\displaystyle y-1=1(x-2)\). Can you rearrange that and put it into standard form?
Jameson
- Admin
- #3
Hello and welcome, Mike!
Let $\displaystyle (x,y)$ be a general point on the line (other than the given point (1,2)). Using the definition of slope, we may write:
$\displaystyle m=\frac{\Delta y}{\Delta x}=\frac{y-1}{x-2}=1$
Note: using the point-slope formula will give us the same result, and more quickly (and once you are familiar with this formula, it is the way to go here), but I wanted to show you an alternate route, and if you think about it, you can see that this is a way to actually derive the point-slope formula.
$\displaystyle \frac{y-1}{x-2}=1$
Now, what should you do first to try to get this into the form:
$\displaystyle Ax+By=C$?
Let $\displaystyle (x,y)$ be a general point on the line (other than the given point (1,2)). Using the definition of slope, we may write:
$\displaystyle m=\frac{\Delta y}{\Delta x}=\frac{y-1}{x-2}=1$
Note: using the point-slope formula will give us the same result, and more quickly (and once you are familiar with this formula, it is the way to go here), but I wanted to show you an alternate route, and if you think about it, you can see that this is a way to actually derive the point-slope formula.
$\displaystyle \frac{y-1}{x-2}=1$
Now, what should you do first to try to get this into the form:
$\displaystyle Ax+By=C$?