# Thread: Need help with linear algebra?

1. ## Need help with linear algebra?

Using Ax+By+C=0 (line 1), determine the point which line 2 touches line 1 where line 2 is perpendicular to line 1, and line 2 touches the origin.

I got some weird result: ( -(A*C)/(B^2+A^2), (A^2 *C -C)/(B^3+A^2 *B) )
Sorry for my computer language.

Gut feeling tells me it's wrong. Anyone concur?

2. Originally Posted by ISeriouslyNeedHelp
Using Ax+By+C=0 (line 1), determine the point which line 2 touches origin and line 1 where line 2 is perpendicular to line 1

I got some weird result: ( -(A*C)/(B^2+A^2), (A^2 *C -C)/(B^3+A^2 *B) )
Sorry for my computer language.

Gut feeling tells me it's wrong. Anyone concur?
x-value is fine ... y-value (meh)

$Ax+By+C=0$

$By = -Ax - C$

$y = -\frac{A}{B}x - \frac{C}{B}$

slope of this line is $m = -\frac{A}{B}$

a normal line has a slope that is the opposite reciprocal ...

$m_p = \frac{B}{A}$

equation of the line with slope $m_p$ that passes thru the origin ...

$y = \frac{B}{A}x$

intersection ... $y = y$

$\frac{B}{A}x = -\frac{A}{B}x - \frac{C}{B}$

multiply every term by $AB$ to clear the fractions ...

$B^2x = -A^2x - AC$

$A^2 x + B^2 x = -AC$

$x = -\frac{AC}{A^2 + B^2}$

$y = \frac{B}{A} \left( -\frac{AC}{A^2 + B^2} \right) = -\frac{BC}{A^2 + B^2}$

3. Thanks. Very big help.