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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).

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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: $$ (am+ bm')x +(an+bn')y+(ar + br') =0$$

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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.

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