Intercept and symmetric equations of a line

Content updated: October 22, 2026

Intercept form uses the intersections with the two coordinate axes, whereas symmetric form uses a point and a direction vector. The two forms have different nonzero-denominator conditions and are not interchangeable.

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1. Intercept form

If a line meets $Ox,Oy$ at $A(a;0)$ and $B(0;b)$, respectively, with $ab\ne0$, then

These two points are distinct and both satisfy the equation, so they determine the line $AB$. The numbers $a,b$ are signed coordinates and need not be positive. From $Ax+By+C=0$, we may take $a=-C/A$ and $b=-C/B$ if $ABC\ne0$. When $C=0$, the line through the origin has no intercept form with right side 1 and two finite nonzero denominators. When $A=0$ or $B=0$, one axis intersection is either not finite or not unique.

2. Symmetric equations and zero denominators

The line through $M_0(x_0;y_0)$ with direction vector $(u;v)$, where $uv\ne0$, has symmetric form

The equality comes from two expressions for the same parameter $t$. If $u=0$ and $v\ne0$, the line has equation $x=x_0$; if $v=0$ and $u\ne0$, its equation is $y=y_0$. Never write a fraction with denominator zero. Parametric equations remain available whenever $(u;v)\ne(0;0)$.

3. Worked examples

Example 1. Two given intercepts

The line through $A(2;0)$ and $B(0;3)$ has equation $x/2+y/3=1$, equivalently $3x+2y-6=0$. Substitution verifies that both given points lie on the line.

Example 2. A negative intercept

Consider $3x-2y+6=0$. Setting $y=0$ gives $x=-2$; setting $x=0$ gives $y=3$. Thus the intercept form is $x/(-2)+y/3=1$. Do not replace $-2$ by the positive length $2$, since that changes the line.

Example 3. Computing intercepts from the correct parametric equations

Let $x=-1+2t$ and $y=2-3t$. At the intersection with $Ox$, the condition $2-3t=0$ gives $t=2/3$, so $x=1/3$. At the intersection with $Oy$, the condition $-1+2t=0$ gives $t=1/2$, so $y=1/2$. Therefore

Substituting the parametric expressions into the general equation gives $3(-1+2t)+2(2-3t)-1=0$ for every $t$, confirming the result.

Example 4. From a point and a direction vector to symmetric form

The line through $M(2;3)$ with direction vector $(-2;1)$ can be written as $(x-2)/(-2)=(y-3)/1$, since both components are nonzero. Cross-multiplication gives $x+2y-8=0$, which contains $M$ and has normal vector $(1;2)$ perpendicular to $(-2;1)$.

Example 5. Converting another pair of parametric equations

For $x=-4+2t$ and $y=3-3t$, we have $t=(x+4)/2=(y-3)/(-3)$. The symmetric form is therefore $(x+4)/2=(y-3)/(-3)$. This line has intercepts $-2,-3$; these are different data from the equations in Example 3.

4. Practice exercises

1. Write the intercept form of $4x+2y-8=0$.
2. For the line through $M(1;-2)$ with direction vector $(0;5)$, can the symmetric form use two fractions? Write a valid equation.
3. Find the intercepts of $x=1+2t$ and $y=-3+t$, where $t\in\mathbb R$.

5. Common mistakes

• Confusing the two denominators in intercept form with the components of a direction vector.
• Replacing a negative coordinate by a positive length.
• Dividing by zero or using intercept form with right side 1 for a line through the origin.
• Combining data from two different parametric representations when finding intersections.

6. Related lessons and resources

• The general equation of a line (English lesson)
• Parametric equations (English lesson)
• The slope of a line (English lesson)

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