General Form, Distances and Families of Lines: Quick Revision

A brief note on General Form, Distances and Families of Lines. Read it before a test. Try each question at the end before you open its answer.

The formulas

• General equation: $Ax + By + C = 0$.
• Slope and intercepts: $-\dfrac{A}{B}$, $-\dfrac{C}{A}$ and $-\dfrac{C}{B}$.
• Normal form: $x\cos\alpha + y\sin\alpha = p$, with $p \ge 0$.
• Point to line: $\dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
• Parallel lines: $\dfrac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
• Family through a meeting point: $L_1 + kL_2 = 0$.
• Parallel to $Ax + By + C = 0$: $Ax + By + k = 0$. Perpendicular: $Bx - Ay + k = 0$.
• A point is on the origin's side of a line when its value has the same sign as $C$.
• Concurrent lines: the meeting point of two lies on the third.
• Shift of origin to $(h, k)$: $X = x - h$ and $Y = y - k$.

A worked example

Find the distance of $(-1, 4)$ from $5x + 12y - 30 = 0$, and the gap between that line and $5x + 12y + 35 = 0$.

Step 1. The distance. $5(-1) + 12(4) - 30 = 13$ and $\sqrt{25 + 144} = 13$, so $d = 1$.

Step 2. The gap. The coefficients already match, so $d = \dfrac{|-30 - 35|}{13} = 5$.

Answer: $1$ and $5$.

Practice with new numbers

A program to try

The program checks both answers of the worked example.

Mistakes to avoid

• The slope as $\dfrac{A}{B}$. It is $-\dfrac{A}{B}$.
• No absolute value. A distance is never negative.
• Parallel lines with different coefficients. Match $A$ and $B$ first.
• Expecting $L_2$ in the family. It is never a member.
• The shift the wrong way. The new coordinate is $x - h$.

Check yourself

1. Find the distance of $(0, 0)$ from $3x + 4y + 10 = 0$.

2. Find the line through $(2, 1)$ perpendicular to $x + 3y - 4 = 0$.

3. Find the fixed point of $(x - y + 1) + k(x + y - 3) = 0$.

4. Are $(1, 1)$ and $(4, 0)$ on the same side of $x + y - 3 = 0$?

5. The origin moves to $(2, 1)$. Write $x - 2y = 0$ in the new axes.