I know I am supposed to ask a specific question, but there's just too many that I would have to ask [it would be like spam] since I missed one week of school because of a family thing and we have an exam this Tuesday, and the teacher's got Mondays off, since it's only 'practice' classes, which means I can't ask her. So, I will just group them here, hoping someone will answer.
Lines
Explicit, Implicit and Segment Line equation
Let's say you have this line equation (implicit form): $x-2y-3=0$
How to convert that (back and forth) into explicit and segment forms.
Common point / Line crossing point
You have two lines: $x-2y-3=0$ and $3x-2y-1=0$
How do you determine where they cross (and if they cross) [This might be a bad example].
Angles between lines
So, taking the two lines from the above example: $x-2y-3=0$ and $3x-2y-1=0$
How would you determine the angle between them (if they're not parallel, that is).
$k$ – the direction coefficient
When given the following line equation: $3x-2y-1=0$. How does one calculate $k$?
Circles
Writing the 'proper' circle equation
I know, the title is a bit… odd, but I will provide an example.
Let's say you're given this circle equation: $x^2+y^2+6x-2y=0$
That has to be transformed into something that resembles: $(x-p)^2+(y-q)^2=r^2$
I would take this approach: $x^2+y^2+6x-2y=0$ / $+3^2-3^2+1-1$
When sorted out you get: $x^2+6x+3^2+y^2-2y+1=8$ which is in fact: $(x+3)^2+(y-1)^2=8$. I hope I'm right! 😛
Defining whether a point is a part of the circle
Let's have we have this circle: $(x+3)^2+(y-1)^2=8$, how would you define whether point $T$ is a part of the circle's 'ring.' I'm going on a limb here, and I'll just point out a thought: Would you just replace the $x$ and $y$ in the equation with the coordinates of $T$?
Tangent
This one's a little tougher (at least I think so). So, you have $(x+3)^2+(y-1)^2=8$ and a point $T(-2,4)$ which can be on or off the circle. Now, I know there's 2 approaches: One if the point is on the circle and the second one if it's off it. So, you have to write a Line equation of the Tangent that goes through that point. I really couldn't figure this one at all I have a vague idea of how to do all the above mentioned, but this one's a bit a total mess.
Circle equation of a circle that touches both of the axis and the circle's centre point lies on a given line
Whew, that took a while to compose… Well, Let's say we have the line $x-2y-6=0$ and we have to determine the centre and the equation of the circle, taking into consideration that the circle touches both axis. The only thing I can gather from that is that $|q|=|p|=r$
Well, I hope someone actually reads and answers this, because I've been writing it for the past hour flipping through the textbook like a madman. And it would save my life.
Best Answer
This is my contribution. Observe that all these equations are equivalent
$$x-2y-3=0\Leftrightarrow y=\frac{1}{2}x-\frac{3}{2}\Leftrightarrow \frac{x}{3}+\frac{y}{-3/2}=1.$$
And the same idea applies to the general equation $Ax+Bx+C=0$.
Lines Common point. Suppose you have two straight lines with equations
$$Ax+By+C=0$$
$$A^{\prime }x+B^{\prime }y+C^{\prime }=0.$$
They are not parallel if and only if $AB^{\prime }-A^{\prime }B\neq 0$ (or equivalently $\frac{B^{\prime }}{B}\neq \frac{A^{\prime }}{A}$)
Since for the lines with equations
$$x-2y-3=0$$
$$3x-2y-1=0$$
we have $\frac{2}{2}\neq \frac{3}{1}$, they cross each other. To find the coordinates $(x,y)$ of the intersecting point you have to solve the system of equations. For instance as follows
$$\left\{ \begin{array}{c} x-2y-3=0 \\ 3x-2y-1=0% \end{array}% \right. \Leftrightarrow \left\{ \begin{array}{c} x-2y-3=0 \\ -3x+2y+1=0% \end{array}% \right. $$
$$\Leftrightarrow \left\{ \begin{array}{c} x-2y-3=0 \\ -2x-2=0% \end{array}% \right. \Leftrightarrow \left\{ \begin{array}{c} -1-2y-3=0 \\ x=-1% \end{array}% \right. \Leftrightarrow \left\{ \begin{array}{c} y=-2 \\ x=-1% \end{array}% \right. $$
The second system results from the first by multiplying the second equation by $-1$. The third, by replacing the second equation by the sum of both equations.
If you solve with matrices
$$% \begin{bmatrix} A & B \\ A^{\prime } & B^{\prime }% \end{bmatrix}% \begin{bmatrix} x \\ y% \end{bmatrix}% =% \begin{bmatrix} -C \\ -C^{\prime }% \end{bmatrix}% $$
the expression $AB^{\prime }-A^{\prime }B$ is the determinant of the matrix formed by the coefficients of $x$ and $y$
$$\begin{bmatrix} A & B \\ A^{\prime } & B^{\prime }% \end{bmatrix}$$
Angles between lines. The smallest angle $0\leq \theta \leq \pi /2$ between two lines is such that
$$\tan \theta =\left\vert \frac{m-m^{\prime }}{1+mm^{\prime }}\right\vert ,$$
where $m$ is the slope of the line with equation $Ax+By+C=0$ and $m^{\prime } $ is the sope of $A^{\prime }x+B^{\prime }y+C^{\prime }=0$.
This result is derived from the trigonometric identity
$$\tan (\alpha -\alpha')=\frac{\tan \alpha-\tan \alpha'}{1+\tan \alpha\cdot\tan \alpha'}$$
where $\alpha,\alpha'$ are the angles of inclination of the first and the second line, respectivelly. And $m=\tan \alpha,m'=\tan \alpha'$.
The other angle between these two lines is $\pi -\theta $.