# Why does $aL_1 + bL_2 =0$ always go through the intersection of $L_1$ and $L_2$

I have this question which asks to explain why $aL_1 + bL_2 =0$, where $L_1$ and $L_2$ are straight lines, will define a straight line through P, the intersection point of $L_1$ and $L_2$.

I can sort of feel the answer but I cannot express it in a cohesive and abstract manner (it stems off the use of the k-method - which finds the line that goes through a given point and the intersection point of 2 other lines, but I just cant really express my reasoning).

If $l_1: mx+ny+r=0$ and $l_2: m'x+n'y+r'=0$ then intersection point $P(x_0,y_0)$ satyfies both of the equations, so:
$mx_0+ny_0+r=0$ and $m'x_0+n'y_0+r'=0$.
Thus $P$ is also on a line (just plug in coordinates of $P$) $a(mx+ny+r) + b(m'x+n'y+r') =0$ which we can rewrite like this: $$(am+ bm')x +(an+bn')y+(ar + br') =0$$
This is a special case of a more general property of zero sets of scalar-valued functions. Suppose we have $$f,g:\mathbb K^n\to\mathbb K$$ and $$S=\{\mathbf p \in \mathbb K^n \mid f(\mathbf p)=0\} \cap \{\mathbf p \in \mathbb K^n \mid g(\mathbf p)=0\}$$, i.e., $$S$$ is the set of common solutions to $$f(\mathbf p)=0$$ and $$g(\mathbf p)=0$$. Then the zero set of every linear combination $$af+bg$$ of the two functions contains $$S$$: clearly, if $$\mathbf p\in S$$, then $$af(\mathbf p)+bg(\mathbf p) = 0$$.
So, in your case, $$aL_1+bL_2$$ passes through $$P$$ and showing that this also represents a straight line is a matter of refactoring the expression.