Standard form of the equation a line

**qleeq** · Nov 25th 2009, 02:01 PM

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 $l$ who lies on points $(2,5)$ and $(7,8)$

Would appreciate an explanation on how to do this, thanks.

**MathTooHard** · Nov 25th 2009, 02:14 PM

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

**qleeq** · Nov 25th 2009, 04:34 PM

Originally Posted by MathTooHard

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

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

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

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

That's where I'm at right now..

**user_5** · Nov 25th 2009, 09:36 PM

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

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

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

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

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

EDIT: Woops, didn't bother checking, just used what he/she already had.

**rowe** · Nov 25th 2009, 11:39 PM

Originally Posted by user_5

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

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

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

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

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

This is not correct.

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

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

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

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

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

In linear form $y= ax + b$ :

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

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

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