Equation of a Line

Equation of a line

There are various ways of defining a line. In the following subsections, a linear equation of the line is given in each case.

Slope–intercept form or Gradient-intercept form

A non-vertical line can be defined by its slope m, and its y-intercept y₀ (the y coordinate of its intersection with the y-axis). In this case its linear equation can be written

$y=mx+y_{0}.$

If, moreover, the line is not horizontal, it can be defined by its slope and its x-intercept x₀. In this case, its equation can be written

$y=m(x-x_{0}),$

or, equivalently,

$y=mx-mx_{0}.$

These forms rely on the habit of considering a non vertical line as the graph of a function. For a line given by an equation

$ax+by+c=0,$

these forms can be easily deduced from the relations

${\begin{aligned}m&=-{\frac {a}{b}},\\x_{0}&=-{\frac {c}{a}},\\y_{0}&=-{\frac {c}{b}}.\end{aligned}}$

Point–slope form or Point-gradient form

A non-vertical line can be defined by its slope m, and the coordinates $x_{1},y_{1}$ of any point of the line. In this case, a linear equation of the line is

$y=y_{1}+m(x-x_{1}),$

or

$y=mx+y_{1}-mx_{1}.$

This equation can also be written

$y-y_{1}=m(x-x_{1})$

for emphasizing that the slope of a line can be computed from the coordinates of any two points.

Intercept form

A line that is not parallel to an axis and does not pass through the origin cuts the axes in two different points. The intercept values x₀ and y₀ of these two points are nonzero, and an equation of the line is

${\frac {x}{x_{0}}}+{\frac {y}{y_{0}}}=1.$

(It is easy to verify that the line defined by this equation has x₀ and y₀ as intercept values).

Two-point form

Given two different points (xTemplate:Sub, yTemplate:Sub) and (xTemplate:Sub, yTemplate:Sub), there is exactly one line that passes through them. There are several ways to write a linear equation of this line.

If xTemplate:Sub ≠ xTemplate:Sub, the slope of the line is ${\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}.$ Thus, a point-slope form is

$y-y_{1}={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}(x-x_{1}).$

By clearing denominators, one gets the equation

$(x_{2}-x_{1})(y-y_{1})-(y_{2}-y_{1})(x-x_{1})=0,$

which is valid also when xTemplate:Sub = xTemplate:Sub (for verifying this, it suffices to verify that the two given points satisfy the equation).

This form is not symmetric in the two given points, but a symmetric form can be obtained by regrouping the constant terms:

$(y_{1}-y_{2})x+(x_{2}-x_{1})y+(x_{1}y_{2}-x_{2}y_{1})=0$

(exchanging the two points changes the sign of the left-hand side of the equation).

Determinant form

The two-point form of the equation of a line can be expressed simply in terms of a determinant. There are two common ways for that.

The equation $(x_{2}-x_{1})(y-y_{1})-(y_{2}-y_{1})(x-x_{1})=0$ is the result of expanding the determinant in the equation

${\begin{vmatrix}x-x_{1}&y-y_{1}\\x_{2}-x_{1}&y_{2}-y_{1}\end{vmatrix}}=0.$

The equation $(y_{1}-y_{2})x+(x_{2}-x_{1})y+(x_{1}y_{2}-x_{2}y_{1})=0$ can be obtained be expanding with respect to its first row the determinant in the equation

${\begin{vmatrix}x&y&1\\x_{1}&y_{1}&1\\x_{2}&y_{2}&1\end{vmatrix}}=0.$

Beside being very simple and mnemonic, this form has the advantage of being a special case of the more general equation of a hyperplane passing through n points in a space of dimension n – 1. These equations rely on the condition of linear dependence of points in a projective space.

Licensing

Content obtained and/or adapted from:

• Linear equation, Wikipedia under a CC BY-SA license

Resources

• Equation of a Line, Khan Academy
• Slope-intercept equation from slope & point, Khan Academy