My Math Forum (http://mymathforum.com/math-forums.php)
-   Algebra (http://mymathforum.com/algebra/)
-   -   A line though two lines' intersection (http://mymathforum.com/algebra/45504-line-though-two-lines-intersection.html)

 Mr Davis 97 July 29th, 2014 06:39 AM

A line though two lines' intersection

I'm doing homework for a math class, and there are problems relating to finding a line that passes though the intersection of two other lines. The general case for this is (apparently) described by the equation
(ax + by + c) + k(dx + ey + f) = 0 where ax + by + c and dx + ey + f are the lines with the point of intersection, and where k is some number. This equation works for finding lines which pass though the intersection of ax + by + c and dx + ey + f, but I do not understand where this equation comes from or why it works. If anybody could explain, I would appreciate it.

 Evgeny.Makarov July 29th, 2014 09:46 AM

This has to be explained in your textbook.

Let $F_1(x,y)=ax + by + c$ and $F_2(x,y)=dx + ey + f$, and let $l_1$ and $l_2$ be lines determined by $F_1(x,y)=0$ and $F_2(x,y)=0$, respectively. Assume $l_1$ and $l_2$ intersect at $P_0(x_0,y_0)$. Then the following fact is true:

All lines passing through $P_0$ are given by equation $\alpha F_1+\beta F_2=0$ for some $\alpha$, $\beta$. (*)

If you take one such equation and $\alpha\ne0$, then you can divide the equation by $\alpha$ without changing the line and get $F_1+kF_2=0$ where $k=\beta/\alpha$. This form is what you are talking about. However, if $\alpha=0$, i.e., if the equation is $F_2=0$, then it cannot be represented in the form $F_1+kF_2=0$ for some $k$ unless $l_1=l_2$, i.e., unless $F_1$ and $F_2$ are proportional. Thus, a more precise statement is:

Every line passing through $P_0$ is either $l_2$ or is determined by equation $F_1+kF_2=0$ for some $k$.

Let's prove (*). Obviously, the line $\alpha F_1+\beta F_2=0$ passes through $P_0$. Conversely, let $l$ pass through $P_0$. Choose some $P'(x',y')\ne P_0$ on $l$; then $P'\notin l_1$ or $P'\notin l_2$ or both, so $F_1(x',y')\ne0$ or $F_2(x',y')\ne0$. Consider
$F_2(x',y')F_1(x,y)+F_1(x',y')F_2(x,y)=0.\qquad(**)$
The coordinates of both $P_0$ and $P'$ satisfy (**), so (**) determines $l$. Strictly speaking, one needs to show that (**) is a first-degree equation and is not degenerate, i.e., does not have the form $0x+0y+0=0$. The fact that $\deg(F_2(x',y')F_1(x,y)+ F_1(x',y')F_2(x,y)) \le1$ is easy since $F_2(x',y')$ and $F_1(x',y')$ are numbers. But if the coefficients of $x$ and $y$ are zero in (**), then $F_2(x',y')(a,b)+F_1(x',y')(d,e)=(0,0)$, i.e., vectors $(a,b)$ and $(d,e)$ are linearly dependent and therefore $l_1$ and $l_2$ are parallel, which is a contradiction.

 All times are GMT -8. The time now is 11:04 PM.