Regarding the equivalence of two forms of family of lines

Question

Suppose we want to find the equation of line passing through the point of intersection of two straight lines $$a_1x+ b_1y+c_1=0$$ and $$a_2x+b_2y+c_2=0$$ There are two ways of doing it.

First way would be to find the common point $(x_1,y_1)$ of this two lines obtained from solving this two equations and then writing the equation of any line passing through this point of intersection as $$y-y_1=m(x-x_1)$$ where m is any quantity whatever.

The second way is, any point common to these two lines will satisfy both of these equations and hence it would satisfy the equation $$a_1x+b_1y+c_1+\lambda(a_2x+b_2y+c_2)=0$$ Thus whatever be the value of $\lambda$ this equation is always satisfied by the point of intersection of the two lines.

Now my question is in the first case $m$ is a parameter by varying which we get the family of straight lines passing through the point of intersection of given two lines and in the second case $\lambda$is the parameter.

But this two family of lines are equivalent thus this two equations should be equivalent. But I have tried to prove the equivalence of this two equations but I have not succeed.

So, how can we show the equivalence of this two equations.

Answer 1

Both methods are incomplete.

Let $a_1=b_1=b_2=1$ and $a_2=0=c_1=c_2$ so that we are dealing with lines $x+y=0$ and $y=0$ and $(0,0)$ is the intersection.

Equation $y=mx$ does not provide the line $x=0$.

Equation $x+y+\lambda y=0$ does not provide the line $y=0$.

This also shows that there is no equivalence because the lines $x=0$ and $y=0$ are distinct.

Answer 2

Refer to drhab's answer, the first way $y−y_1=m(x−x_1)$ is incomplete for all lines passing through $(x_1,y_1)$ (here we don't consider the degenerated case that the two lines $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ are parallel) . However, as drhab also commented, we can change the form as:

$$n(y−y_1)=m(x−x_1)\quad\text{(1)}$$

Analogously, we can change the second way as:

$$\kappa(a_1x+b_1y+c_1)+\lambda(a_2x+b_2y+c_2)=0\quad\text{(2)}$$

In fact, these two ways are equivalent because both (1) and (2) can be represented as:

$$ax+by+c=0\quad\text{(3)}$$

In the first way:

$$\begin{cases}a=m\\b=-n\\c=ny_1-mx_1=n\dfrac{\det\left[\begin{matrix}c_1&a_1\\c_2&a_2\end{matrix}\right]}{\det\left[\begin{matrix}a_1&b_1\\a_2&b_2\end{matrix}\right]}-m\dfrac{\det\left[\begin{matrix}b_1&c_1\\b_2&c_2\end{matrix}\right]}{\det\left[\begin{matrix}a_1&b_1\\a_2&b_2\end{matrix}\right]}\end{cases}\quad\text{(1')}$$

In the second way:

$$\begin{cases}a={\kappa}a_1+{\lambda}a_2\\b={\kappa}b_1+{\lambda}b_2\\c={\kappa}c_1+{\lambda}c_2\end{cases}\quad\text{(2')}$$

Both (1') and (2') can be represented as:

$$\det\left[\begin{matrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a&b&c\end{matrix}\right]=0\quad\text{(3')}$$