Family of lines and intersection problems

Two non-parallel lines meet at exactly one point. Three or more lines that all pass through a common point are concurrent. A clever way to describe all lines through a given point is the family of lines: any line through the intersection of two given lines can be written as $L_1 + k L_2 = 0$ for some real $k$ .

Intersection of two lines

To find the intersection of $A_1 x + B_1 y + C_1 = 0$ and $A_2 x + B_2 y + C_2 = 0$ , solve the linear system. Cramer's rule gives: $x = \frac{B_1 C_2 - B_2 C_1}{A_1 B_2 - A_2 B_1}, \quad y = \frac{C_1 A_2 - C_2 A_1}{A_1 B_2 - A_2 B_1}.$

The denominator $A_1 B_2 - A_2 B_1$ is zero iff the lines are parallel (no unique intersection).

Family of lines through an intersection

Theorem. If $L_1 = 0$ and $L_2 = 0$ are two non-parallel lines intersecting at $P$ , then every line through $P$ has the form $L_1 + k L_2 = 0$ for some scalar $k$ (and the limit $k \to \infty$ gives $L_2 = 0$ ).

The reason: both $L_1$ and $L_2$ vanish at $P$ , so the combination vanishes too , meaning $P$ lies on $L_1 + k L_2 = 0$ . Conversely, any line through $P$ can be written this way by suitably choosing $k$ .

Strategy. Given that a line passes through the intersection of two known lines and satisfies one more condition (slope, distance, perpendicularity, $\dots$ ), use the family form $L_1 + k L_2 = 0$ , impose the extra condition, solve for $k$ .

Concurrency

Three lines $L_1 = 0, L_2 = 0, L_3 = 0$ are concurrent iff $\det \begin{pmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{pmatrix} = 0.$

Equivalently, $L_3$ can be written as a linear combination $\alpha L_1 + \beta L_2$ .

Worked examples

Example 1. Find the intersection of $2 x + 3 y - 7 = 0$ and $x - y + 1 = 0$ .

From the second: $x = y - 1$ . Substitute: $2(y - 1) + 3 y - 7 = 0 \Rightarrow 5 y = 9 \Rightarrow y = 9/5, x = 4/5$ . Intersection: $(4/5, 9/5)$ .

Example 2. Find the line through the intersection of $x + y - 3 = 0$ and $2 x - y + 1 = 0$ , passing through $(1, 4)$ .

Family: $(x + y - 3) + k(2 x - y + 1) = 0$ . Plug in $(1, 4)$ : $(1 + 4 - 3) + k(2 - 4 + 1) = 0 \Rightarrow 2 + k(-1) = 0 \Rightarrow k = 2$ .

Line: $x + y - 3 + 2(2 x - y + 1) = 0 \Rightarrow 5 x - y - 1 = 0$ .

Example 3. Show that $3 x + 4 y - 10 = 0$ , $x - y - 1 = 0$ , and $5 x + 6 y - 16 = 0$ are concurrent.

Solve the first two: $3 x + 4 y = 10$ and $x - y = 1 \Rightarrow x = y + 1 \Rightarrow 3(y + 1) + 4 y = 10 \Rightarrow 7 y = 7 \Rightarrow y = 1, x = 2$ . Check the third: $5 \cdot 2 + 6 \cdot 1 - 16 = 0$ ✓.

Example 4. Find the line through the intersection of $x + 2 y - 5 = 0$ and $3 x + y - 10 = 0$ , perpendicular to $4 x + 3 y - 7 = 0$ .

Intersection: solve to get $x = 3, y = 1$ . Required slope: $4/3$ (negative reciprocal of $-4/3$ ).

Equation: $y - 1 = (4/3)(x - 3) \Rightarrow 4 x - 3 y - 9 = 0$ .

Example 5. For what value of $k$ are the lines $k x + 2 y + 3 = 0$ , $x + k y + 6 = 0$ , and $3 x + 2 y + 5 = 0$ concurrent?

Determinant: $\det\begin{pmatrix} k & 2 & 3 \\ 1 & k & 6 \\ 3 & 2 & 5 \end{pmatrix} = 0$ .

Expand: $k(5k - 12) - 2(5 - 18) + 3(2 - 3k) = 5 k^2 - 12 k + 26 + 6 - 9 k = 5 k^2 - 21 k + 32 = 0$ .

Discriminant: $441 - 640 < 0$ . No real $k$ .

Try it yourself

Find the intersection of $3 x + 2 y = 7$ and $x - y = 1$ .
Find the line through $(2, 3)$ and the intersection of $x + y - 4 = 0$ and $2 x - y - 1 = 0$ .
Find the line through the intersection of $x + 2 y - 1 = 0$ and $x - y + 4 = 0$ perpendicular to $x + y - 1 = 0$ .
Show that $(1, 0)$ , $(0, 1)$ , and the intersection of $x + y - 2 = 0$ and $2 x - y + 1 = 0$ are collinear.
Are the lines $x - y - 1 = 0$ , $2 x - y - 3 = 0$ , $3 x - y - 5 = 0$ concurrent? Find the common point.
Find $k$ if the lines $x + y - 1 = 0$ , $2 x + 3 y - 5 = 0$ , $x + k y - 4 = 0$ are concurrent.
Find the line through the intersection of $3 x + y - 5 = 0$ and $5 x - y - 7 = 0$ which makes equal intercepts on the axes.
Find the equation of the line through $(2, -1)$ and through the intersection of $x - 2 y + 1 = 0$ and $x + 3 y - 1 = 0$ .
Find the foot of the perpendicular from $(1, 2)$ to the line $4 x + 3 y - 25 = 0$ .
Find the image of $(2, 3)$ in the line $y = x$ .
Three sides of a triangle have equations $L_1, L_2, L_3$ . Show that the medians are concurrent (without using calculus). (Hint: use the family of lines.)
Find the line through the intersection of $x + 2 y - 3 = 0$ and $3 x + 4 y - 7 = 0$ at distance $1$ from origin.

Pitfalls / Tricks

For the family form, $k$ can be any real; including $k \to \infty$ gives $L_2 = 0$ .
Three lines may be parallel without being concurrent (parallel ≠ concurrent). Concurrency means they meet at a single point.
When solving the linear system, check whether the lines are parallel first.
Insight. The family of lines through a point is a powerful technique: it reduces "find the line through point $P$ satisfying condition $X$ " to a single-variable equation in $k$ .