# three questions on analytic geometry and matrices

1. the lines $x-2y=4$ and $6x+ay=8$ are perpendicular. Calculate the value of $a$.

2. prove that the matrix $\pmatrix{\cos\theta& \sin\theta\\ -\sin\theta & \cos\theta}$ is never singular (equal to zero).

3. the points $(2,6)$ , $(-4,21)$ , $(7,k)$ are collinear. Determine the value of $k$.

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What is higher homework? Does this fall under the tag (homework)? –  Did Feb 20 '13 at 18:44
Are you sure that $cos0$, $sin0$ are what you are supposed to have in that matrix? Not $\cos x$ and $\sin x$? Or maybe it was $\cos \theta$ and $\sin \theta$? –  MJD Feb 20 '13 at 18:46
I replaced $0$ in the matrix with $\theta$, bcause "never singular" would seem to imply you wanted a variable not a constant –  Thomas Andrews Feb 20 '13 at 18:47
@bailey: Usually when people ask homework questions here, we like them to also say what they have tried, and how far they got, and what part they are stuck on. –  MJD Feb 20 '13 at 18:48
i have tried but i was off from school at that point and i wasnt sure how to do it sorry MJD. –  bailey Feb 20 '13 at 18:51

Hints:

For (1), two lines are perpendicular if the product of their slopes is $-1$.

For (2), that matrix is a rotation matrix, rotating the plane $\theta$. Show that the expected inverse is an inverse. (Alternatively, if you know that matrices with non-zero determinant are non-singular, you can show by computing the determinant.)

Aside: Non-singular is not equivalent to "not zero." There are non-zero matrices that are singular. Non-singular is equivalent to the determinant being non-zero, however.

For (3), do you know how to determine the equation of the line going through $(2,6)$ and $(-4,21)$?

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Writing lines in explicit form $$x-2y=4\iff y=\frac{1}{2}x-2$$ and $$6x+ay=8\iff y=-\frac{6}{a}x+\frac{8}{a}$$ lines $y=k_1x+n_1$ and $y=k_2x+n_2$ are perpendicular if $k_1=-\frac{1}{k_2}$ in our case $$\frac{1}{2}=-\frac{1}{-6/a}\Rightarrow a=3$$

Points $(2,6)$, $(-4,21)$, $(7,k)$ are colineare if the triangle that they form has area $0$ or if $$|2(21-k)+(-4)(k-6)+7(6-21)|=0\Rightarrow k=-\frac{87}{6}$$

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