Forms of Equation of a Straight Line

9.3 Various Forms of the Equation of a Line

Every straight line in the plane admits an algebraic equation of first degree in \(x\) and \(y\). Depending on the data given (a slope, a point, two points, intercepts, a perpendicular from origin, etc.) we use different convenient forms. We derive each one and illustrate with examples.

9.3.1 Horizontal and vertical lines

A horizontal line at signed distance \(a\) from the \(x\)-axis has equation \(y=a\) (with \(a>0\) if above the axis, \(a<0\) below). A vertical line at signed distance \(b\) from the \(y\)-axis has equation \(x=b\).

9.3.2 Point–slope form

Let \(L\) be a line with slope \(m\) passing through a known point \(P_0(x_0,y_0)\). For any other point \(P(x,y)\) on \(L\) (with \(x\neq x_0\)) we have \(\dfrac{y-y_0}{x-x_0}=m\), which rearranges to:

\[\boxed{\;y-y_0=m(x-x_0)\;}\quad\text{(point–slope form)}\]

9.3.3 Two-point form

If \(L\) passes through \(P_1(x_1,y_1)\) and \(P_2(x_2,y_2)\), its slope is \(m=\dfrac{y_2-y_1}{x_2-x_1}\). Substituting in the point–slope form at \(P_1\):

\[\boxed{\;y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\;}\]

9.3.4 Slope–intercept form

If line \(L\) has slope \(m\) and cuts the \(y\)-axis at \((0,c)\), applying point–slope with \((x_0,y_0)=(0,c)\):

\[\boxed{\;y=mx+c\;}\]

Here \(c\) is called the y-interceptⁱ. If instead \(L\) makes \(x\)-intercept \(d\), one obtains \(y=m(x-d)\).

9.3.5 Intercept form

Suppose the line cuts the \(x\)-axis at \((a,0)\) and the \(y\)-axis at \((0,b)\), with \(a\neq 0,b\neq 0\). Using the two-point form:

\(y-0=\dfrac{b-0}{0-a}(x-a)\ \Rightarrow\ -ay=b x-ab\ \Rightarrow\ \dfrac{x}{a}+\dfrac{y}{b}=1.\)

\[\boxed{\;\dfrac{x}{a}+\dfrac{y}{b}=1\;}\quad\text{(intercept form)}\]

9.3.6 Normal form

Let \(p\) be the perpendicular distance from the origin to line \(L\), and let \(\omega\) be the angle that this perpendicular (drawn from origin onto \(L\)) makes with the positive \(x\)-axis. If \(A(a,0)\) and \(B(0,b)\) are the intercepts of \(L\), then from right-triangle geometry at the foot of perpendicular \(N\):

\(\cos\omega=\dfrac{p}{a},\ \sin\omega=\dfrac{p}{b}\Rightarrow a=\dfrac{p}{\cos\omega},\ b=\dfrac{p}{\sin\omega}.\) Substituting in the intercept form:

\[\boxed{\;x\cos\omega+y\sin\omega=p\;}\quad (p\ge 0,\ 0\le\omega<2\pi)\]

9.3.7 General equation of a line

From \(Ax+By+C=0\) we can recover all previous forms:

• Slope–intercept: \(y=-\dfrac{A}{B}x-\dfrac{C}{B}\).
• Intercept: \(\dfrac{x}{-C/A}+\dfrac{y}{-C/B}=1\) (provided \(C\neq 0\)).
• Normal: divide by \(\pm\sqrt{A^2+B^2}\) choosing the sign so the right side is positive: \(\dfrac{A}{\pm\sqrt{A^2+B^2}}x+\dfrac{B}{\pm\sqrt{A^2+B^2}}y=\dfrac{-C}{\pm\sqrt{A^2+B^2}}\).

9.4 Miscellaneous Examples