Hi There!Was hoping for some help.

Set up an equation for the line (2,3) which is perpendicular to the line (2, -1) and (-1.-3). Don't understand this at all. Can someone please help me?!

Thx!

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- Dec 12th 2009, 12:20 PMWalkmanPerpendicular line equation!
**Hi There!**Was hoping for some help.

Set up an equation for the line (2,3) which is perpendicular to the line (2, -1) and (-1.-3). Don't understand this at all. Can someone please help me?!

Thx! - Dec 12th 2009, 12:25 PMe^(i*pi)
Use the equation of a straight line: $\displaystyle y=m_1x+c$

where $\displaystyle m_1 = \frac{y_2-y_1}{x_2-x_1}$

You have the two points for a line so you can work this out.

For two lines to be perpendicular their gradients $\displaystyle m_1$, $\displaystyle m_2$ must equal -1 and so $\displaystyle m_2 = -\frac{1}{m_1}$

Now $\displaystyle m_2 $ is known use the point (2,3) to get an equation of the form $\displaystyle y = mx+c$ - Dec 12th 2009, 12:43 PMWalkman
- Dec 12th 2009, 12:53 PMe^(i*pi)
$\displaystyle m_1 = \frac{-3-(-1)}{-1-2} = \frac{-2}{-3} = \frac{2}{3}$

Using the point (2,-1): $\displaystyle -1 = \frac{2}{3}(2)+c \: \rightarrow \: c = -\frac{7}{3}$

line 1: $\displaystyle y = \frac{2x-7}{3}$

Indeed you are right (Rofl)

As you now know the gradient of the above line is $\displaystyle \frac{2}{3}$ so $\displaystyle m_2 = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}$

Are you ok to do the same as you did before but with the point (2,3) and the gradient -1.5 - Dec 12th 2009, 01:32 PMWalkman
- Dec 13th 2009, 09:45 PMWalkman
Just as i thought. Can't get it right...

Keep getting wierd answers all the time...

help please..? - Dec 14th 2009, 04:15 AMmr fantastic