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My teacher said that whenever you're given a line, say $ax+by+c=0$,

and a linear relation in $a$, $b$ and $c$, say $2a + 3b+c=0$,

these two sets of equations represent a family of lines {in this case passing through $(2,3)$}.

However, he did not elaborate on it. The books that I have at home only mentions the form $L_1 + k L_2$ = 0, where $L_1$ and $L_2$ are two non-parallel lines and $k$ is a parameter. I know that when we eliminate $c$ from the equation of the line by substituting its value from the linear relation, we will get the family of lines in the form $a(x-2) +b (y-3) = 0$ and $k =a/b$.

But can someone intuitively explain why the former happens... (like in the case of the latter $L_1$ and $L_2$ both satisfy that point, so $0 + k 0 = 0 $, something like this which helps me get the 'feel' for it). P.S. Sorry I couldn't find a better word for feel.

Ernie060
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prarabdh shivhare
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  • As a side note, observe that $L_1+kL_2=0$ doesn’t describe all of the lines passing through the intersection of $L_1$ and $L_2$: there’s no value of $k$ that produces the equation $L_2=0$, so $L_2$ itself is omitted from the family. Be aware of this when you use this in the course of solving a problem. You either have to check $L_2$ as a separate case or use a linear combination $aL_1+bL_2=0$. – amd Apr 19 '20 at 19:53

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Observe that $2a+3b+c=0$ is the equation that results from substituting the coordinates of the point $(2,3)$ into the equation $ax+by+c=0$ of a line. So, if you have three numbers $a$, $b$ and $c$ that satisfy this linear constraint, then the equation of a line that has those three numbers for its coefficients is satisfied by $(2,3)$. To put it another way, the linear relation $2a+3b+c=0$ expresses the fact that $(2,3)$ lies on the line $ax+by+c=0$.

amd
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