# Thread: equation of a straight line through A(-1,3) perpendicular to y=1/3x-1?

1. ## equation of a straight line through A(-1,3) perpendicular to y=1/3x-1?

hi,

im not too sure how to work this out:

equation of a straight line through A(-1,3) perpendicular to y=1/3x-1

i require some help on this one step at a time please.

thanks!

2. ## Re: equation of a straight line through A(-1,3) perpendicular to y=1/3x-1?

Originally Posted by andyboy179
hi,

im not too sure how to work this out:

equation of a straight line through A(-1,3) perpendicular to y=1/3x-1

i require some help on this one step at a time please.

thanks!
1. Two lines with the slopes $\displaystyle m_1$ and $\displaystyle m_2$ are perpendicular if the slopes satisfy the equation

$\displaystyle m_1 \cdot m_2 = -1$

So the slope of the perpendicular line is (-3).

2. Use the point-slope-formula of a straight line to determine the equation of the line in question:

$\displaystyle y-y_1 = m(x-x_1)$

where $\displaystyle x_1, y_1$ are the coordinates of the point which is located on the line:

$\displaystyle y-3=(-3)(x-(-1))~\implies~y=-3x$

3. ## Re: equation of a straight line through A(-1,3) perpendicular to y=1/3x-1?

Originally Posted by andyboy179
equation of a straight line through A(-1,3) perpendicular to y=1/3x-1

So applying that to this we get $\displaystyle y=-3x+~?$

4. ## Re: equation of a straight line through A(-1,3) perpendicular to y=1/3x-1?

Hi andyboy179.
The slope of the line perpendicular to y=1/3 x -1 is -3because the product of the slopes must be -1. Use the point slope formula to find the new line

y-3/x+1 = -3 finish it

bjh

y=x+1?

6. ## Re: equation of a straight line through A(-1,3) perpendicular to y=1/3x-1?

Originally Posted by Plato

So applying that to this we get $\displaystyle y=-3x+~?$
but if you look on that link i don't understand it!!!

7. ## Re: equation of a straight line through A(-1,3) perpendicular to y=1/3x-1?

actually would it be y=-3+4?

8. ## Re: equation of a straight line through A(-1,3) perpendicular to y=1/3x-1?

You have the line: y=-3x+?, now you have also a point A(-1,3) on the straight line so you have to fill in the coordinates in the general equation:
3=-3(-1)+? ...