General Form of Equation of a Line

5.4 General Form of Equation of a Line

Every line's equation can be written in the form

called the general form. For example, $y=3x+2$ rewrites as $3x-y+2=0$, and $\dfrac{x}{2}+\dfrac{y}{3}=1$ rewrites as $3x+2y-6=0$.

From the general form:

• the slope is $-\dfrac{a}{b}$ (provided $b\ne0$);
• the X-intercept is $-\dfrac{c}{a}$ (provided $a\ne0$);
• the Y-intercept is $-\dfrac{c}{b}$ (provided $b\ne0$).

Remark. If $a=0$, the line is parallel to the X-axis and has no X-intercept; if $b=0$, the line is parallel to the Y-axis and has no Y-intercept.

Worked Example 1. Find the slope and intercepts of (a) $x+y+10=0$, (b) $2x+y+30=0$, (c) $x+3y-15=0$.

1. $a=1,b=1,c=10$: slope $=-1$, X-int $=-10$, Y-int $=-10$.
2. $a=2,b=1,c=30$: slope $=-2$, X-int $=-15$, Y-int $=-30$.
3. $a=1,b=3,c=-15$: slope $=-\tfrac13$, X-int $=15$, Y-int $=5$. Worked Example 2. Find the acute angle between (a) $12x-4y=5$ and $4x+2y=7$; (b) $y=2x+3$ and $y=3x+7$.

(a) Slopes are $m_1=3$, $m_2=-2$: $\tan\theta=\left|\dfrac{3-(-2)}{1+3(-2)}\right|=\left|\dfrac{5}{-5}\right|=1 \Rightarrow \theta=45°$.

(b) Slopes are $m_1=2$, $m_2=3$: $\tan\theta=\left|\dfrac{2-3}{1+6}\right|=\dfrac17 \Rightarrow \theta=\tan^{-1}\!\left(\dfrac17\right)$.

Worked Example 3. Find the acute angle between $y-\sqrt3x+1=0$ and $\sqrt3y-x+7=0$. Slopes: from the first, $y=\sqrt3x-1$, so $m_1=\sqrt3$; from the second, $y=\dfrac{x}{\sqrt3}-\dfrac{7}{\sqrt3}$, so $m_2=\dfrac{1}{\sqrt3}$. Then $\tan\theta=\left|\dfrac{\sqrt3-\frac{1}{\sqrt3}}{1+\sqrt3\cdot\frac{1}{\sqrt3}}\right|=\left|\dfrac{\frac{2}{\sqrt3}}{2}\right|=\dfrac{1}{\sqrt3} \Rightarrow \theta=30°$.

Worked Example 4. Show the following pairs are perpendicular: (a) $2x-4y=5$ and $2x+y=17$; (b) $y=2x+23$ and $2x+4y=27$.

(a) $m_1=\tfrac12$, $m_2=-2$: $m_1m_2=-1$ ✓. (b) $m_1=2$, $m_2=-\tfrac12$: $m_1m_2=-1$ ✓.

Worked Example 5. Find the lines through the origin making $45°$ with $3x-y=6$ (slope $3$). Let $m$ be the unknown slope: $\tan45°=\left|\dfrac{m-3}{1+3m}\right|=1$, so $\dfrac{m-3}{1+3m}=\pm1$. Taking $+1$: $m-3=1+3m \Rightarrow -2m=4 \Rightarrow m=-2$. Taking $-1$: $m-3=-1-3m \Rightarrow 4m=2 \Rightarrow m=\tfrac12$. The two required lines through the origin are $y=-2x$ and $y=\tfrac12x$, i.e. $2x+y=0$ and $x-2y=0$.

Worked Example 6. A line is parallel to $2x+y=7$ (slope $-2$) and passes through the origin: $y=-2x$, i.e. $2x+y=0$.

Worked Example 7. A line is parallel to $x+3y=9$ (slope $-\tfrac13$) and passes through $A(2,7)$: $y-7=-\tfrac13(x-2) \Rightarrow 3y-21=-x+2 \Rightarrow x+3y-23=0$. …