Why Intersecting Lines Give a Unique Solution

From Two Lines to One Ordered Pair

A point lies at the intersection of two lines precisely when its coordinates satisfy both line equations. Thus, finding an intersection means finding an ordered pair \((x,y)\) that solves a two-equation system. When the lines are nonparallel, elimination does more than produce a candidate point: it determines both coordinates, and those values are forced for any point satisfying the equations.

Work over \(\mathbb{R}\), and keep the variable order \((x,y)\) throughout. Write the equations of two lines as

where \(a,b,c,d,e,f\in\mathbb{R}\), and \((a,b)\ne(0,0)\), \((d,e)\ne(0,0)\). These conditions ensure that each equation defines a line rather than an equation with no \(x\)- or \(y\)-coefficient. A common point is an ordered pair \((x,y)\) that makes both equations true.

The coefficient pairs determine the directions of the lines. For the first line, a displacement \((u,v)\) between two of its points must satisfy \(au+bv=0\). The nonzero vector \((-b,a)\) satisfies that equation and gives a direction for the line. The second line has direction vector \((-e,d)\). Two lines are nonparallel when their direction vectors are not scalar multiples of one another.

Here is why that coefficient test matches the geometric definition. The direction vector \((-b,a)\) is obtained from the coefficient pair \((a,b)\) by the same rule used to obtain \((-e,d)\) from \((d,e)\). This rule is reversible: from a direction vector \((u,v)=(-b,a)\), the coefficient pair is \((a,b)=(v,-u)\). So the two direction vectors are scalar multiples exactly when their corresponding coefficient pairs are scalar multiples.

Proposition. If \((a,b)\ne(0,0)\) and \((d,e)\ne(0,0)\), then \((a,b)\) and \((d,e)\) are scalar multiples exactly when \(ae-bd=0\).

Proof. If \((d,e)=k(a,b)\) for a scalar \(k\), then \(d=ka\) and \(e=kb\). Substitution gives \(ae-bd=a(kb)-b(ka)=0\). Conversely, suppose \(ae-bd=0\). If \(a\ne0\), set \(k=d/a\). The equation \(ae=bd\) gives \(e=bd/a=kb\), while \(d=ka\), so \((d,e)=k(a,b)\). If \(a=0\), then \(b\ne0\). The equality \(ae-bd=0\) becomes \(-bd=0\), and therefore \(d=0\). Set \(k=e/b\); then \((d,e)=(0,e)=k(0,b)=k(a,b)\). This proves both directions, including the case \(a=0\). \(\square\)

It follows that nonparallel lines have \(\Delta\ne0\): their direction vectors are not scalar multiples, so their coefficient pairs are not scalar multiples, and the proposition gives the nonzero condition. This condition is what makes the elimination steps below possible.

How Elimination Determines the Coordinates

Assume now that the lines are nonparallel, so \(\Delta\ne0\). Multiply the first equation by \(e\), the second by \(b\), and subtract the second resulting equation from the first. The \(y\)-terms cancel:

Since \(ae-bd=\Delta\ne0\), division by \(\Delta\) is valid and determines \(x\). To determine \(y\), multiply the second equation by \(a\), the first by \(d\), and subtract the latter from the former. This time the \(x\)-terms cancel:

Again dividing by the nonzero number \(\Delta\) is valid. These computations give the candidate coordinates:

Proof. Because the lines are nonparallel, \(\Delta\ne0\). Define \(x=(ce-bf)/\Delta\) and \(y=(af-dc)/\Delta\). Substituting these values into the left side of the first equation gives

For the second equation, substitution gives

Thus the displayed pair satisfies both equations, so a common point exists. To prove uniqueness, take any ordered pair satisfying both equations. Multiplying and subtracting the equations as above forces \(\Delta x=ce-bf\) and \(\Delta y=af-dc\). Since \(\Delta\ne0\), these equations force \(x=(ce-bf)/\Delta\) and \(y=(af-dc)/\Delta\). Every common point therefore has the displayed coordinates. Existence and uniqueness are both proved. \(\square\)

The elimination steps are justified because they combine equalities that are already true for any common solution. They show that every such solution must have the same coordinates. The substitution calculation then confirms that those forced coordinates really do satisfy both original equations.

Worked Examples

Why the Nonzero Condition Matters

The uniqueness argument depends on dividing by \(\Delta\). If \(\Delta=0\), those particular elimination equations do not determine \(x\) and \(y\) by division. That is not a permission to divide anyway, and it is not evidence that a unique solution has been found. The theorem concerns the nonparallel case, where the coefficient condition is nonzero. In a calculation, check that condition before using the coordinate formulas.

Common mistake: dividing by a zero coefficient. For the line \(x=4\), a student might try to isolate \(y\) by treating its equation as \(x+0y=4\) and dividing by the coefficient of \(y\). That would require division by \(0\), which is not defined. The equation already specifies \(x=4\); it places no restriction on \(y\) by itself. When paired with \(y=-2\), use the two equations together: they give \(x=4\) and \(y=-2\), and substitution verifies the common point. More generally, the coordinate formulas divide by \(\Delta\), not by an individual coefficient that may be zero.

Geometry and algebra describe the same fact in two ways. Geometrically, nonparallel directions allow at most one point to lie on both lines. Algebraically, the nonzero value of \(\Delta\) lets elimination force a single value for each coordinate. The substitution check establishes that the forced pair is actually on both lines. Keeping these roles separate—condition, calculation, and verification—makes the uniqueness conclusion precise.