# Standard form of the equation a line

• Nov 25th 2009, 02:01 PM
qleeq
Standard form of the equation a line
Alright, I know I've done this before but I can't remember how to set it up to save my life.

Write in standard form the equation of line $\displaystyle l$ who lies on points $\displaystyle (2,5)$ and $\displaystyle (7,8)$

Would appreciate an explanation on how to do this, thanks.
• Nov 25th 2009, 02:14 PM
MathTooHard
Given two points and the slope between them, you can write the line in point slope form: $\displaystyle y-y_1=m(x-x_1)$, where $\displaystyle m$ is the slope. The slope is rise over run, which is $\displaystyle \frac{8-5}{7-2}=\frac{3}{5}$. Now, we choose any given point and plug it into the equation. Let's choose $\displaystyle (2,5)$, and we get $\displaystyle y-5=\frac{3}{5}(x-2)$. From there you can manipulate the equation to standard form: $\displaystyle ax+by=c$, where $\displaystyle a, b, c$ are integers.
• Nov 25th 2009, 04:34 PM
qleeq
Quote:

Originally Posted by MathTooHard
Given two points and the slope between them, you can write the line in point slope form: $\displaystyle y-y_1=m(x-x_1)$, where $\displaystyle m$ is the slope. The slope is rise over run, which is $\displaystyle \frac{8-5}{7-2}=\frac{3}{5}$. Now, we choose any given point and plug it into the equation. Let's choose $\displaystyle (2,5)$, and we get $\displaystyle y-5=\frac{3}{5}(x-2)$. From there you can manipulate the equation to standard form: $\displaystyle ax+by=c$, where $\displaystyle a, b, c$ are integers.

So just to make sure..
$\displaystyle y-5=\frac{3}{5}(x-2)$

What do I do with the $\displaystyle y-5$ part?

$\displaystyle y-5=\frac{3}{5}x+1\frac{1}{5}$

That's where I'm at right now..
• Nov 25th 2009, 09:36 PM
user_5
Add 5 to both sides, by changing 5 into a number with a denominator of 5

$\displaystyle y-5+5=\frac{3}{5}x+\frac{6}{5}+5$

$\displaystyle y=\frac{3}{5}x+\frac{6}{5}+\frac{25}{5}$

$\displaystyle y=\frac{3}{5}x+\frac{31}{5}$

$\displaystyle y=\frac{3}{5}x+6\frac{1}{5}$

• Nov 25th 2009, 11:39 PM
rowe
Quote:

Originally Posted by user_5
Add 5 to both sides, by changing 5 into a number with a denominator of 5

$\displaystyle y-5+5=\frac{3}{5}x+\frac{6}{5}+5$

$\displaystyle y=\frac{3}{5}x+\frac{6}{5}+\frac{25}{5}$

$\displaystyle y=\frac{3}{5}x+\frac{31}{5}$

$\displaystyle y=\frac{3}{5}x+6\frac{1}{5}$

This is not correct.

$\displaystyle y-5 = \frac{3}{5}x-\frac{6}{5}$

$\displaystyle y = \frac{3x-6}{5}+5$

$\displaystyle y = \frac{3x-6}{5}+\frac{25}{5}$

$\displaystyle y = \frac{3x-6+25}{5}$

$\displaystyle y = \frac{3x+19}{5}$

In linear form $\displaystyle y= ax + b$:

$\displaystyle y = \frac{3}{5}x+\frac{19}{5}$

Plug your two example pairs (2,5) and (7,8) to verify this is correct.