Question 48814
Hi,

Your line passes through the points (-3, -4) and (0, 5). To get from the first point to the second, the line must go right 3 units and up 9 units. If you only went right one unit then you would go up three units.

So the gradient vector for the line is (1, 3) this means for every unit you go right, you go three up.

The perpendicular line has a gradient vector (1, m) where m is the gradient. This means that for every unit you go right, you go m units up.

Now these two gradient vectors must be perpendicular because the lines are perpendicular. Two vectors are perpendicular when their dot product is 0. So if we set the dot product to 0 we should get an equation to calculate m.

*[tex \left(\begin{array}{c}1\\3\end{array}\right)\cdot\left(\begin{array}{c}1\\m\end{array}\right)=0]

This leads to the equation *[tex 1+3m=0] So *[tex m=-1/3].

The simple rule I learnt many years ago in school was that if *[tex g] is the gradient of a line, then a perpendicular line will have gradient *[tex -1/g] This can actually be proved using the above technique on a general line.

Hope that helps,
Kev