# equation of a line - help

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• November 13th 2012, 03:50 AM
chg
equation of a line - help
Hi,

I'm stuck on a problem and I need help... I'm supposed to find the general equation of a line (Ax+By+C=0) from two points. I did a couple of these problems and everything went well but not for this one... so here goes:

points: (-sqrt(2)/2, sqrt(2)/2), (1/2, sqrt(3)/2)

We are given the answer but I can't figure out how to get there... Here is the answer:

(sqrt(3) - sqrt(2))x - (1 + sqrt(2))y + (sqrt(2) + sqrt(6)) / 2 = 0

Anyone can help me with the steps to get to this answer?

This is my first post so go easy on me if there is a better way to format equations (just tell me how to).

Thanks!
• November 13th 2012, 05:38 AM
darthjavier
Re: equation of a line - help
Hi

You have these points: $\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right); \left(\frac{1}{2},\;\frac{}{\sqrt{3}}{2}\right)$

The slope of the line is: $\mathrm{m}=\dfrac{\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}}{\frac{1}{2}+\frac{\sqrt{2}}{2} }$

Now consider a general point $\mathrm{(x,\;y)}$ of this line, which must have the same slope:

$\mathrm{m=\dfrac{\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}}{\frac{1}{2}+\frac{\sqrt{2}}{2} }=\dfrac{y-\frac{\sqrt{2}}{2}}{x+\frac{\sqrt{2}}{2}}}$

Now you got it (Cool)

Greetings :)
• November 13th 2012, 06:54 AM
chg
Re: equation of a line - help
Hi and thanks for your answer.

I'm pretty new to maths so what might seem obvious for you is still a mystery for me :)

I did understand the slope formula but I can't see why the y difference (sqrt(3) - sqrt(2)) transfers to the A variable in Ax+By+C=0 (same thing for the x difference that becomes B...) Also, how you get the C variable from these differences is beyond me.

Sorry if this is too much explanation to have me understand something simple but I just don't get it :(

Thanks
• November 13th 2012, 07:14 AM
darthjavier
Re: equation of a line - help
$\mathrm{m}$ is the slope

$\mathrm{m=\dfrac{\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}}{\frac{1}{2}+\frac{\sqrt{2}}{2} }=\dfrac{y-\frac{\sqrt{2}}{2}}{x+\frac{\sqrt{2}}{2}}}$

or

$\mathrm{m=\dfrac{\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}}{\frac{1}{2}+\frac{\sqrt{2}}{2} }=\dfrac{y-\frac{\sqrt{3}}{2}}{x-\frac{1}{2}}}$

Now multiply in each side

$\\\mathrm{\dfrac{\frac{\sqrt{3}}{\rlap{/}2}-\frac{\sqrt{2}}{\rlap{/}2}}{\frac{1}{\rlap{/}2}+\frac{\sqrt{2}}{\rlap{/}2}}=\dfrac{y-\frac{\sqrt{3}}{2}}{x-\frac{1}{2}}}}\\\mathrm{x(\sqrt{3}-\sqrt{2})-\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}=y(1+\sqrt{ 2})-\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{6}}{2}}\\\mathrm{x(\sqrt{3}-\sqrt{2})-y(1+\sqrt{2})+\dfrac{\sqrt{2}+\sqrt{6}}{2}}=0$

$\mathrm{A=\sqrt{3}-\sqrt{2}}$

$\mathrm{B=-(1+\sqrt{2})}$

$\mathrm{C=\dfrac{\sqrt{2}+\sqrt{6}}{2}}$

:)
• November 13th 2012, 07:32 AM
HallsofIvy
Re: equation of a line - help
You give the equation as Ax+ By= C. Do you realize those numbers are not "unique"? You could multiply the entire equation by any constant. You could, for example, Divide the entire equation by C to get (A/C)x+ (B/C)y= 1. Rewriting that as A'x+ B'y= 1 (A'= A/C and B'= B/C), putting $x= -\sqrt{2}/2$, $y= \sqrt{2}/2$ we have $\frac{\sqrt{2}}{2}A'+ \frac{\sqrt{2}}{2}B'= 1$ and multiplying both sides by $\frac{2}{\sqrt{2}}= \sqrt{2}$, $A'+ B'= \sqrt{2}$.

Now put x= 1/2, $y= \sqrt{3}{2}$ in the equation: $A'/2+ \sqrt{3}B'/2= 1$ so that $A'+ \sqrt{3}B'= 2$. Now, subtracting the previous equation from that, the A' is eliminated: $(\sqrt{3}- \sqrt{2})B'= 1$. Solve that for B', then put that into either of the previous equations.
• November 13th 2012, 07:52 AM
chg
Re: equation of a line - help
Thank you very much guys, that was very helpful (a lot more than my exercise book). Looks like I still have some basics to grasp, but with some help like this, I'll get there eventually :)

Cheers!