Perpendicular line equation!

**Walkman** · December 12th 2009, 12:20 PM

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!

**e^(i*pi)** · December 12th 2009, 12:25 PM

Originally Posted by Walkman

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!

Use the equation of a straight line: $y=m_1x+c$

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

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

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

Now $m_2$ is known use the point (2,3) to get an equation of the form $y = mx+c$

**Walkman** · December 12th 2009, 12:43 PM

Originally Posted by e^(i*pi)

Use the equation of a straight line: [YOUTUBE]y=m_1x+c[/YOUTUBE]

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

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

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

Now $m_2$ is known use the point (2,3) to get an equation of the form $y = mx+c$

Hm. Maybe i took the wrong way solving this. But just before you replied i got the equation fot the line between the two given points.

y = (2 x)/3-7/3.

dont know if it's correct though...

**e^(i*pi)** · December 12th 2009, 12:53 PM

Originally Posted by Walkman

Hm. Maybe i took the wrong way solving this. But just before you replied i got the equation fot the line between the two given points.

y = (2 x)/3-7/3.

dont know if it's correct though...

$m_1 = \frac{-3-(-1)}{-1-2} = \frac{-2}{-3} = \frac{2}{3}$

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

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

Indeed you are right

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

Are you ok to do the same as you did before but with the point (2,3) and the gradient -1.5

**Walkman** · December 12th 2009, 01:32 PM

Originally Posted by e^(i*pi)

$m_1 = \frac{-3-(-1)}{-1-2} = \frac{-2}{-3} = \frac{2}{3}$

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

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

Indeed you are right

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

Are you ok to do the same as you did before but with the point (2,3) and the gradient -1.5

Here's where my problem always come in. I never solves the perpendicular line. I know how to do it, but for some reason i can't do it... I'll try..

**Walkman** · December 13th 2009, 09:45 PM

Just as i thought. Can't get it right...

Keep getting wierd answers all the time...

help please..?

**mr fantastic** · December 14th 2009, 04:15 AM

Originally Posted by Walkman

Just as i thought. Can't get it right...

Keep getting wierd answers all the time...

help please..?

You have been shown excatly how to do this question and given all the information required for doing it. Your troubles will be caused by poor algebra. So please show all your working so that your mistake(s) can be explained to you.

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