Let $L_1$ be a line, $a_1X + b_1Y+ c_1 =0$, $L_2$ be a line, $a_2X + b_2Y+ c_2= 0$.
Then prove that the equation of line(s) passing through the intersection of these two line is of type $\mathbf{L_1 + KL_2 = 0}$.
We can prove it by multiply two equations with different constants and then adding them..., this proof didn't worked for me,therefore i need a best proof, prove it directly or give me some awesome hints.
Let $C(x_0,y_0)$ be the common point $L_1 \cap L_2$. Thus
$$\begin{cases}a_1x_0 + b_1y_0+ c_1 =0\\a_2x_0 + b_2y_0+ c_2 =0\end{cases}$$
By difference with the initial equations, we obtain the new equivalent equations:
$$\tag{1}\begin{cases}a_1(x-x_0) + b_1(y-y_0)=0 \ \ (L_1)\\a_2(x-x_0) + b_2 (y-y_0)=0 \ \ (L_2)\end{cases} \ \ \implies \ \underbrace{(a_1+Ka_2)(x-x_0) + (b_1+Kb_2)(y-y_0)=0}_{\text{line} (L_1+KL_2)}$$
As any line passing through C has (for fixed coefficients $u$ and $v$) the following equation:
$$\tag{2}u(x-x_0) + v(y-y_0)=0,$$ it suffices to check that there can exist a value of $K$ such that
$$\tag{3}\frac{a_1+Ka_2}{u}= \frac{b_1+Kb_2}{v}$$
(equations (1) and (2) should not necessarily be the same, but they must have proportionnal coefficients).
(3) is a first degree equation that has a single solution, except for a special case where $a_2v-b_2u=0$ (i.e., line $(L_2)$ : see my comment just after the question.)
Remarks:
a) In (3) we have assumed $u\neq0$ and $v\neq0$. This little technical difficulty is removed if one uses, instead of fractions, a determinant equal to zero.
b) What we have done is equivalent to an origin shift.
Here are some hints...
Hint 1:
At some point you need to consider the intersection point between $L_1$ and $L_2$... Let's say the intersectino is $P$, what can you say of $X_P$, $Y_P$ and the various coefficients?
Hint 2:
Consider any line $L$ of equation $aX+bY+c=0$ and assume it belongs to the family, what can you say about its coefficients?
EDIT removed hints 3 and 4 since they pointed to a wrong proof
