Equation for the family of lines that passes through $3y-5x-10=0$ and $3y-\frac{x}{3}-\frac{5}{3}=0$ 
Find an equation for the family of lines that passes through the intersection of $3y-5x-10=0$ and $3y-\frac{x}{3}-\frac{5}{3}=0$

 A: I think it's $$\displaystyle k \left(3y-5x-10 \right)+m \left(3y-\frac{x}{3}-\frac{5}{3} \right)=0$$ where $k$ and $m$ are not all zero.
A: Let us denote by $L = L(a, b, c)$ the line $ax + by + c = 0$.
$(a, b, c)$ can be considered a 3-dimensional vector $v$ and all lines
can be represented by the 3-dimensinal vector space $R^3$.
The bundle of 
 lines $L(a, b, c)$ passing through the intersection $(i, j)$ of the 
given lines $L_1 = L(a_1, b_1, c_1)$ and $L_2 = L(a_2, b_2, c_2)$ satisfy 
$$a i + b j + c = 0$$
So, the vectors $v$ corresponding to this bundle lie in a 2-dimensional
subspace $R^2$. $v_1 = (a_1, b_1, c_1)$ and $v_2 = (a_2, b_2, c_2)$ are also in this
subspace because they satisfy the above restriction.
Moreover, they are linearly independent (because the 2 lines cross)
and hence they span this subspace. Therefore $v$ must be a linear
combination of $v_1$ and $v_2$:
$$v = \lambda_1 v_1 + \lambda_2 v_2$$.
So, the bundle lines are given by
$$L(\lambda_1 a_1 + \lambda_2 a_1, \lambda_1 b_1 + \lambda_2 a_2, \lambda_1 c_1 + \lambda_2 c_2)$$
$$ = \lambda_1 L_1 + \lambda_2 L_2$$
because $L$ is a linear form. The line equation of these lines is 
$$\lambda_1 L_1 + \lambda_2 L_2 = 0$$
(as used without proof in another answer). 
Now, we should divide by $\lambda_1$ to get the family of bundle lines
parametrized by one parameter $\lambda$: 
$$L_1 + \lambda L_2 = 0$$
A: Find the intersection of $3x - 5y - 10 = 3x - \frac{x}{3} - \frac{5}{3}$, it should be a point  in the plane $(a,b)$, then shift the origin to $(a,b)$.
For example, $x=y$ becomes $(x-a)=(y-b)$, to make it have an arbitrary slope $m$, just scale the left hand side of the equation to get $(x-a)m=(y-b)$, so the family of all lines that go through the point $(a,b)$ (except the vertical one) is given by $\{(x,y) \text{ } | \text{ } (x-a)m=(y-b), m\in\mathbb{R} \}$.
In your example, $(a,b) = \left(\frac{5}{14},-\frac{25}{14}\right)$, here are some of the lines in that family, sampled by selecting $m$ values from $[-5,5]$.
$\left(\frac{5}{14},-\frac{25}{14}\right)$" />
