Expert: Josh - 8/21/2008

Question
Hi.  I have a question in my homework, but I don't know how to do it, can you help? The question is:

Find the value of k if the given pairs of lines are perpendicular;

x+ky=3
x-y=1

Answer
Hi Joanna,

We can always arrange two straight line equations in the form of
y=mx+b and y=nx+c, where "m and n" represent the gradient, "b and c" represent the y-intercept of the first and second lines respectively.

The main thing to remember is that "m*n=-1" if the two lines are perpendicular.

Here, x+ky=3 can be turned into y=-(1/k)x+3/k.
The second equation x-y=1 may be rearranged as y=x-1.

Comparing y=-(1/k)x+3/k to y=mx+b, and y=x-1 to y=nx+c, the gradient of the first line m=-1/k, while the gradient of the second line is given by n=1.

Since m*n=-1, we need (-1/k)*1=-1. The solution is k=1.
Of course, if you prefer, you can start from m=-1/n. Setting m=-1/k, this leads to -1/k = -1/1, so k=1.

Once you are familiar with the result m*n=-1, you can find the answer often by inspection.