Find where two lines meet, test whether three lines are concurrent, and use the family $l_1 + \lambda l_2 = 0$ to write lines through an intersection point without finding the point.
Two lines in a plane either are parallel (gradients equal, no common point), coincide, or meet at exactly one point. To find the point, solve the two equations together (substitution or elimination); the solution $(x, y)$ is the point of intersection. If the elimination produces a false statement such as $0 = 5$, the lines are parallel.
Example: Find the intersection of $2x + 3y = 13$ and $x - y = -1$. From the second, $x = y - 1$. Then $2(y - 1) + 3y = 13$, so $5y = 15$, $y = 3$ and $x = 2$. The lines meet at $(2, 3)$.
Example: Find the area of the triangle formed by the lines $x + y = 4$, $x - y = 0$ and the $x$-axis. Intersections: $x + y = 4$ and $x - y = 0$ give $(2, 2)$; $x + y = 4$ with $y = 0$ gives $(4, 0)$; $x - y = 0$ with $y = 0$ gives $(0, 0)$. The base is $4$ and the height is $2$, so the area is $\dfrac12 \cdot 4 \cdot 2 = 4$.
Three lines are concurrent if they have a common point. To test: find the intersection of two of them and check that it satisfies the third. A shortcut: the lines $a_ix + b_iy + c_i = 0$ ($i = 1, 2, 3$) are concurrent when $$\begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} = 0$$ (provided no two of the lines are parallel).
Example: Show that $x + y - 3 = 0$, $2x - y = 0$ and $3x - y - 1 = 0$ are concurrent. The first two meet where $y = 2x$ and $3x = 3$, that is at $(1, 2)$. In the third line, $3(1) - 2 - 1 = 0$, so it passes through $(1, 2)$ as well.
Let $l_1 \equiv a_1x + b_1y + c_1$ and $l_2 \equiv a_2x + b_2y + c_2$ meet at $P(x_0, y_0)$. Then $l_1(P) = 0$ and $l_2(P) = 0$, so for any number $\lambda$ $$l_1 + \lambda l_2 = 0 \quad\text{i.e.}\quad (a_1 + \lambda a_2)x + (b_1 + \lambda b_2)y + (c_1 + \lambda c_2) = 0$$ is satisfied by $P$. It is a linear equation in $x$ and $y$, so it is a line, and it passes through $P$ for every $\lambda$. Different values of $\lambda$ give different lines through $P$; this is a family of lines with parameter $\lambda$.
The power of this idea: you can write down a line through the intersection of two lines without solving for the intersection point. Then fix $\lambda$ with one more condition.
Example: Find the line through the intersection of $x + 2y - 3 = 0$ and $2x - y + 4 = 0$ that also passes through the origin. The family is $(x + 2y - 3) + \lambda(2x - y + 4) = 0$. At $(0, 0)$: $-3 + 4\lambda = 0$, so $\lambda = \dfrac34$. Multiply the equation by $4$: $4(x + 2y - 3) + 3(2x - y + 4) = 0$, which simplifies to $10x + 5y = 0$, that is $2x + y = 0$.
Example: Find the line through the intersection of $x + y - 4 = 0$ and $2x - y + 1 = 0$ that is perpendicular to $3x - 2y - 5 = 0$. The family is $(1 + 2\lambda)x + (1 - \lambda)y + (\lambda - 4) = 0$. A line perpendicular to $3x - 2y - 5 = 0$ has $a_1a_2 + b_1b_2 = 0$: $3(1 + 2\lambda) - 2(1 - \lambda) = 0$, so $1 + 8\lambda = 0$ and $\lambda = -\dfrac18$. Substituting: $\dfrac34 x + \dfrac98 y - \dfrac{33}{8} = 0$; multiply by $8$ to get $6x + 9y - 33 = 0$, that is $2x + 3y - 11 = 0$. (The intersection point is $(1, 3)$, and $2 + 9 - 11 = 0$.)
Fixed point. If a line has a parameter $k$ in the equation, rearrange it as $l_1 + k, l_2 = 0$. Then it always passes through the intersection of $l_1 = 0$ and $l_2 = 0$.
Example: Show that $(2 + k)x + (1 - k)y - (5 + k) = 0$ passes through a fixed point for all $k$ and find it. Group by $k$: $(2x + y - 5) + k(x - y - 1) = 0$. The fixed point solves $2x + y = 5$ and $x - y = 1$, giving $3x = 6$, so the point is $(2, 1)$.
Exam tip: Whenever a question says "the line through the point of intersection of ... and ...", use $l_1 + \lambda l_2 = 0$ immediately; it avoids solving fractions. State clearly that the family excludes $l_2 = 0$ and, if the condition is "parallel to the $y$-axis" or similar, check that the coefficient of $y$ is zero rather than dividing by it. In "show that the line passes through a fixed point" parts, collect the terms in the parameter first, equate each bracket to zero and solve. Concurrency parts usually come first, with the family used in the next part.