# 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 $3y-\frac{x}{3}-\frac{5}{3}=0$

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.

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 $\lambda$: $$L_1 + \lambda L_2 = 0$$

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 $$[-5,5]$$. $\left(\frac{5}{14},-\frac{25}{14}\right)$" />