1. Perpendicular 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!

2. 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 $y = mx+c$

3. Originally Posted by e^(i*pi)

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...

4. 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

5. 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..

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

Keep getting wierd answers all the time...