Relative positions and intersections of two lines in the plane

2026-10-22 huynhvietkhanh Comments: 0

Classify intersecting, parallel, and coincident lines by their equation system or determinant, including zero coefficients.

High schoolPlane Coordinate Geometry

Content updated: October 22, 2026

Two lines in a plane may intersect, be parallel, or coincide. Solving their two equations both finds any intersection and determines their relative position. There are no skew lines in Oxy.

Learning goals: classify two lines and find their intersection, using determinants without dividing by coefficients that may be zero.

Prerequisites: systems of two linear equations in two unknowns and general equations of lines.

On this page: Concepts · Method · Examples · Practice · Further learning

1. The number of solutions of the two-line system

Let $d_i:a_ix+b_iy+c_i=0$ ($i=1,2$), with $(a_i;b_i)\ne(0;0)$ for each line. The coordinates of their common points solve the system

$$\begin{cases}a_1x+b_1y=-c_1,\\a_2x+b_2y=-c_2.\end{cases}$$

Exactly one solution: the lines intersect at the point with those coordinates.
No solution: the lines are distinct and parallel.
Infinitely many solutions: the lines coincide.

Merely saying that the system “has a solution” does not establish that the lines meet at exactly one point, since they may coincide. Before using this classification, also check that both equations really define lines.

2. Determinants and coefficient checks

Set $D=a_1b_2-a_2b_1$. If $D\ne0$, the system has the unique solution

$$x=\frac{b_1c_2-b_2c_1}{D},\qquad y=\frac{a_2c_1-a_1c_2}{D}.$$

If $D=0$, the normal vectors are parallel. Choose a point $M$ on $d_1$: if $M\in d_2$, the lines coincide; otherwise they are distinct and parallel. Equivalently, check $a_1c_2-a_2c_1$ and $b_1c_2-b_2c_1$. If both vanish, the lines coincide; if either is nonzero, they are parallel. This avoids ratios with zero denominators.

3. Worked examples

Example 1. Solving the correct system and checking the result

Consider $d_1:2x-5y+3=0$ and $d_2:5x+2y-3=0$. We have $D=2\cdot2-5(-5)=29\ne0$. Therefore

$$x=\frac{(-5)(-3)-2\cdot3}{29}=\frac9{29},\qquad y=\frac{5\cdot3-2(-3)}{29}=\frac{21}{29}.$$

The intersection is $M(9/29;21/29)$. Check: $2\cdot9/29-5\cdot21/29+3=0$ and $5\cdot9/29+2\cdot21/29-3=0$. The point $(1;-1)$ satisfies only the second equation and is therefore not a common point.

Example 2. Parallel or coincident?

For $x-2y+1=0$ and $2x-4y+3=0$, the normal vectors are parallel. Multiplying the first equation by 2 gives $2x-4y+2=0$, contradicting the second equation if a common point existed. The lines are distinct and parallel. If the second line is replaced by $-3x+6y-3=0$, its entire equation is $-3$ times the first, so the lines coincide.

Example 3. Zero coefficients cause no difficulty

The lines $x-2=0$ and $y+1=0$ intersect at $(2;-1)$. Their normal vectors are $(1;0)$ and $(0;1)$, both nonzero, and $D=1$. Ratios involving their zero components are unnecessary.

4. Practice exercises

Classify $y=x$ and $y=x+2$.
Classify $2x+3y-6=0$ and $4x+6y-12=0$.
Find the intersection of $x=1$ and $y=-2$.

Show answers and hints

Both lines have slope 1 but different y-intercepts, so they are distinct and parallel.
The second equation is twice the first, so the lines coincide.
The unique intersection is $(1;-2)$.

5. Common mistakes

Copying a coefficient of $y$ incorrectly or failing to change the sign of $c_i$ when moving it to the right side.
Accepting a solution without substituting it into both equations.
Concluding that two lines meet at one point merely because the system has a solution.
Using coefficient ratios when a denominator is zero.

6. Related lessons and resources

Open the original exercise resource on Scribd. This lesson includes practice exercises with answers.