you don't have to tell me how it's derived if you prefer not to. i can ask my teacherQuote:

Originally Posted byThePerfectHacker

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- Apr 18th 2006, 07:17 PMc_323_hQuote:

Originally Posted by**ThePerfectHacker**

- Apr 19th 2006, 09:08 AMThePerfectHackerQuote:

Originally Posted by**c_323_h**

$\displaystyle Ax+By+C=0$ a non-vertical non-horizonal line.

Thus, $\displaystyle y=-\frac{A}{B}x-\frac{C}{B}$.

This, is important: to find the distance to point $\displaystyle (x_0.y_0)$ we need to draw a perpendicular line from that point to our line. Then, find the point where it intersects. And then finally use the distance formula to find the distance between these two points.

1)Find the slope of perpendicular line. That is easy since the perpendicular line has a negative reciprocal we have its slope is $\displaystyle \frac{B}{A}$.

2)Find the equation of this perpendicular line. Using the slope-intercept formula we have, (because this perpendicualr line passes through $\displaystyle (x_0,y_0)$)

$\displaystyle y-y_0=\frac{B}{A}(x-x_0)$

Thus,

$\displaystyle Ay-Bx=D$ where $\displaystyle D=Ay_0-Bx_0$

3)Find intersection of these two lines. The original line has equation,

$\displaystyle By+Ax=-C$ and the perpendicular line $\displaystyle Ay-Bx=D$.

Thus the problem reduces to finding the solution of these two systems,

$\displaystyle \left\{ \begin{array}{c}By+Ax=-C\\Ay-Bx=D$

The determinant of these two systems is,

$\displaystyle \left| \begin{array}{cc} B&A\\A&-B\end{array} \right|=-A^2-B^2$

Using Cramer's rule,

$\displaystyle y=\frac{\left| \begin{array}{cc} -C&A\\D&-B\end{array} \right|}{-A^2-B^2}=\frac{AD-BC}{A^2+B^2}=\frac{A^2y_0-ABx_0-BC}{A^2+B^2}$

$\displaystyle x=\frac{\left| \begin{array}{cc} B&-C\\A&D\end{array} \right|}{-A^2-B^2}=\frac{BD+AC}{-A^2-B^2}=\frac{B^2x_0-ABy_0-AC}{A^2+B^2}$

This is the point of intersection between the lines.

4)Find the distance. The distance between point $\displaystyle (x_0,y_0)$ and $\displaystyle \left(\frac{B^2x_0-ABy_0-AC}{A^2+B^2},\frac{A^2y_0-ABx_0-BC}{A^2+B^2} \right)$

Which is,

$\displaystyle \sqrt{\frac{\left( \frac{B^2x_0-ABy_0-AC}{A^2+B^2}-x_0 \right)^2+\left( \frac{A^2y_0-ABx_0-BC}{A^2+B^2}-y_0 \right)^2}{}}$

**GET READY FOR THE NIGHTMARE**

I am going to work with each one seperately. First the first squared sum and then the other one (too big for LaTex :eek: ):

$\displaystyle

\left( \frac{B^2x_0-ABy_0-AC}{A^2+B^2}-x_0 \right)^2=\left(\frac{-ABy_0-AC-A^2x_0}{A^2+B^2} \right)^2

$

The second squared sum under the radical is,

$\displaystyle

\left( \frac{A^2y_0-ABx_0-BC}{A^2+B^2}-y_0 \right)^2=\left( \frac{-ABx_0-BC-B^2y_0}{A^2+B^2} \right)^2

$

Open parantheses of each one and add them up, I found it helpful using the identity $\displaystyle (a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$

-------------------------------------------------------------------------

$\displaystyle

A^2B^2y_0^2+A^2C^2+A^4x_0^2+2A^2BCy_0+2A^3Bx_0y_0+ 2A^3Cx_0$$\displaystyle +A^2B^2x_0^2+B^2C^2+B^4y_0^2+2AB^2Cx_0+2AB^3x_0y_0 +2B^3Cy_0}$

That is numerator the denominator is of course $\displaystyle (A^2+B^2)^2$

Working with the numerator notice we can group terms and factor as,

$\displaystyle

(A^2+B^2)C^2+(A^2+B^2)B^2y_0^2+(A^2+B^2)A^2x_0^2$$\displaystyle +(A^2+B^2)2BCy_0+(A^2+B^2)2ACx_0+(A^2+B^2)2ABx_0y_ 0$

Factoring again,

$\displaystyle (A^2+B^2)$$\displaystyle (C^2+B^2y_0^2+A^2x_0^2+2BCy_0+2ACx_0+2ABx_0y_0)$

But notice the expression,

$\displaystyle (C^2+B^2y_0^2+A^2x_0^2+2BCy_0+2ACx_0+2ABx_0y_0)$=$\displaystyle (Ax_0+By_0+C)^2$ by the identity I mentioned above.

Thus, after all this we finally reach,

$\displaystyle

\sqrt{\frac{\left( \frac{B^2x_0-ABy_0-AC}{A^2+B^2}-x_0 \right)^2+\left( \frac{A^2y_0-ABx_0-BC}{A^2+B^2}-y_0 \right)^2}{}}

$=$\displaystyle \sqrt{\frac{(A^2+B^2)(Ax_0+By_0+C)^2}{(A^2+B^2)^2} }$

Notice, the common factors kill eachother $\displaystyle (A^2+B^2)$ thus,

$\displaystyle \sqrt{\frac{(Ax_0+By_0+C)^2}{A^2+B^2}}$

Since both the numerator and denominator are positive we can split the radical as,

$\displaystyle \frac{\sqrt{(Ax_0+By_0+C)^2}}{\sqrt{A^2+B^2}}$

Remember by the propert of absolute value you have $\displaystyle \sqrt{n^2}=|n|$ thus,

$\displaystyle \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$

$\displaystyle \mathbb{Q}.\mathbb{E}.\mathbb{D}$ - Apr 19th 2006, 12:15 PMc_323_h
:) wow! thanks! you seem very knowledgable in math. you should try to solve (prove or disprove) riemann hypothesis and share it with MHF :D

- Apr 19th 2006, 12:35 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

- Apr 20th 2006, 11:50 AMtopsquark
Yes, he is very good isn't he?

He knows almost enough to start learning Physics. :D

-Dan

(Flinches) I'm waiting with great amounts of anxiety for the repost!