It’s been a while since my last math post. It seems that my previous assertion correlating the quality of question with the period of final exams could not have been further from the truth. It now seems quite likely that the occurrence of a few “good” questions on Yahoo Answers in a short period of time a few weeks back was more of a fluke rather than the norm.

**Problem**: Find the standard matrix for the linear transformation which reflects points in the x-y plane across the line .

**Solution**: To find the matrix representing a given linear transformation all we need to do is to figure out where the basis vectors, i.e.

and

are mapped under the transformation.

This is easy to see from the fact that since and form a basis of any element

can be written as

for some .

Now let

be a linear transformation. Then, by definition

Thus, the image of any element under a linear transformation is completely determined by the image of the basis of the transformation’s domain.

Let’s find the image of under a reflection across the line given by . First, we need to find a line perpendicular to that passed through the point .

In general, a line perpendicular to is given by , i.e in our case, . To find the one particular line with the same slope, that passes through , we need to simply plug in the values of and into the line equation and add a constant value so that the equality will hold. Then we have

Thus, the line perpendicular to that passes through is given by

Now let’s find the point at which our two lines and intersect. This is done easily by solving the two line equation simultaneously:

Equating and (and multiplying by 6 for simplicity) gives

which then gives

Thus, our two lines intersect at

Now, let be the reflection of across . Since and are symmetric across (because that is how the reflection is actually constructed) it follows that

so

Similarly as above, we find the image of under a reflection across the line given by . The line perpendicular to that passed through the point is given by

Equating and (and multiplying by 6 again) gives

and so

Therefore, and intersect at

and the image of under the reflection across is given by

which yields

Finally, the matrix for the linear transformation which reflects points in the x-y plane across the line is given by

It turns out (although I’m not going to show to derivation here) that in general, the matrix for a linear transformation which reflects points in the x-y plane across an arbitrary line is given by

where is the angle that the line makes with the positive x-axis, i.e

if

and

if