Distance from a point to a line

Core Understanding: What ARE We Finding?

We want the perpendicular distance from a point $P(x_1, y_1)$ to a line $L: ax + by + c = 0$.

Key insight: Among all possible paths from $P$ to any point on line $L$, the perpendicular path is the shortest. Why? Because any slanted path forms the hypotenuse of a right triangle, which is always longer than the perpendicular leg.

Deriving the Formula from First Principles

Method 1: Area of Triangle Approach

Let's take a line $ax + by + c = 0$ and a point $P(x_1, y_1)$ not on the line.

Step 1: Find TWO points on the line.

• Set $x = 0$: we get $by + c = 0 \Rightarrow y = -\frac{c}{b}$ (if $b \neq 0$), giving point $A(0, -\frac{c}{b})$
• Set $y = 0$: we get $ax + c = 0 \Rightarrow x = -\frac{c}{a}$ (if $a \neq 0$), giving point $B(-\frac{c}{a}, 0)$

Step 2: Calculate the distance $AB$ (base of our triangle). $AB = \sqrt{\left(-\frac{c}{a}\right)^2 + \left(-\frac{c}{b}\right)^2} = \sqrt{\frac{c^2}{a^2} + \frac{c^2}{b^2}} = |c|\sqrt{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{|c|}{\sqrt{a^2 + b^2}} \cdot \frac{\sqrt{a^2+b^2}}{\sqrt{a^2b^2}} \cdot \sqrt{a^2+b^2}$

Wait, let me recalculate: $AB = \sqrt{\frac{c^2}{a^2} + \frac{c^2}{b^2}} = |c|\sqrt{\frac{b^2 + a^2}{a^2b^2}} = \frac{|c|\sqrt{a^2+b^2}}{|ab|}$

Step 3: Calculate area of triangle $PAB$ using the coordinate formula: $\text{Area} = \frac{1}{2}|x_1(y_A - y_B) + x_A(y_B - y_1) + x_B(y_1 - y_A)|$ $= \frac{1}{2}\left|x_1\left(-\frac{c}{b} - 0\right) + 0\left(0 - y_1\right) + \left(-\frac{c}{a}\right)\left(y_1 + \frac{c}{b}\right)\right|$ $= \frac{1}{2}\left|-\frac{cx_1}{b} - \frac{cy_1}{a} - \frac{c^2}{ab}\right| = \frac{|c|}{2|ab|}\left|ax_1 + by_1 + c\right|$

Step 4: Use the relationship Area = $\frac{1}{2} \times \text{base} \times \text{height}$.

If $d$ is the perpendicular distance (height), then: $\frac{|c|}{2|ab|}|ax_1 + by_1 + c| = \frac{1}{2} \times \frac{|c|\sqrt{a^2+b^2}}{|ab|} \times d$

Simplifying: $|ax_1 + by_1 + c| = \sqrt{a^2+b^2} \cdot d$

Method 2: Perpendicular from Point (Vector Approach)

Why this works: The perpendicular distance is the component of any position vector in the direction of the normal to the line.

The line $ax + by + c = 0$ has a normal vector $\vec{n} = (a, b)$.

Take any point $Q(x_0, y_0)$ on the line (so $ax_0 + by_0 + c = 0$).

The vector from $Q$ to $P$ is $\vec{QP} = (x_1 - x_0, y_1 - y_0)$.

The perpendicular distance is the projection of $\vec{QP}$ onto the unit normal: $d = \left|\vec{QP} \cdot \frac{\vec{n}}{|\vec{n}|}\right| = \frac{|a(x_1-x_0) + b(y_1-y_0)|}{\sqrt{a^2+b^2}}$

Since $ax_0 + by_0 = -c$: $d = \frac{|ax_1 + by_1 - (ax_0 + by_0)|}{\sqrt{a^2+b^2}} = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}$

Worked Examples

Solution: Here $a = 3, b = 4, c = -10$, and $(x_1, y_1) = (3, 4)$.

$d = \frac{|3(3) + 4(4) + (-10)|}{\sqrt{3^2 + 4^2}} = \frac{|9 + 16 - 10|}{\sqrt{9 + 16}} = \frac{|15|}{\sqrt{25}} = \frac{15}{5} = 3$

Why each step?

• We substitute directly into the formula.
• $\sqrt{3^2 + 4^2} = 5$ is a3-45 Pythagorean triple—recognize these for speed!
• The distance is 3 units.

Solution: Here $a = 1, b = -2, c = 4$, and $(x_1, y_1) = (-1, 2)$.

$d = \frac{|1(-1) + (-2)(2) + 4|}{\sqrt{1^2 + (-2)^2}} = \frac{|-1 - 4 + 4|}{\sqrt{1 + 4}} = \frac{|-1|}{\sqrt{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$

Why each step?

• The numerator becomes negative ($-1$), but the absolute value makes it positive.
• We rationalize $\frac{1}{\sqrt{5}}$ by multiplying by $\frac{\sqrt{5}}{\sqrt{5}}$.
• Distance is $\frac{\sqrt{5}}{5} \approx 0.447$ units.

Solution: Step 1: Find the equation of the line. Slope $m = \frac{5-1}{4-1} = \frac{4}{3}$.

Using point-slope form: $y - 1 = \frac{4}{3}(x - 1)$ $3(y - 1) = 4(x - 1)$ $3y - 3 = 4x - 4$ $4x - 3y - 1 = 0$

Step 2: Apply the distance formula. Here $a = 4, b = -3, c = -1$, and $(x_1, y_1) = (2, -3)$.

$d = \frac{|4(2) + (-3)(-3) + (-1)|}{\sqrt{4^2 + (-3)^2}} = \frac{|8 + 9 - 1|}{\sqrt{16 + 9}} = \frac{16}{\sqrt{25}} = \frac{16}{5}$

Why each step?

• We must convert to $ax + by + c = 0$ form first—slope-intercept won't work directly.
• The distance is $\frac{16}{5} = 3.2$ units.

Common Mistakes

Why it feels right: You might think distance can be negative (like a "signed distance"), but distance is always non-negative.

Steel-man: In some contexts (like determining which side of a line a point is on), the signed distance is useful. But when asked for "the distance," we want the magnitude.

Fix: Always include $|\cdots|$ around the numerator.

Why it feels right: $y = mx + c$ is the most familiar form.

The issue: The formula requires the form $ax + by + c = 0$. If you use $y = 2x + 3$, you must rewrite it as $-2x + y - 3 = 0$ (or $2x - y + 3 = 0$).

Fix: Always convert to $ax + by + c = 0$ first. Remember: $y = mx + k \Rightarrow mx - y + k = 0$.

Why it feels right: You might forget the square root, thinking it's just $a^2 + b^2$.

Fix: The denominator is $\sqrt{a^2 + b^2}$, not $a^2 + b^2$. This comes from the magnitude of the normal vector $(a, b)$.

Special Cases

When the point lies ON the line: $ax_1 + by_1 + c = 0$, so $d = 0$. ✓

Parallel lines: To find the distance between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$, pick ANY point on one line and find its distance to the other: $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$

Deep Connections

• Connections:
  • Equation of a Line: Must convert to $ax+by+c=0$ form
  • Perpendicular Lines: The shortest distance is along the perpendicular
  • Projection of Vectors: Method2 uses vector projection
  • Area of Triangle Using Coordinates: Method 1 derives from area
  • Distance Formula: Foundation for all distance calculations
  • Normal Vector to a Line: $(a, b)$ is perpendicular to $ax+by+c=0$
  • Linear Programming: Distance constraints use this formula
  • Least Squares Regression: Minimizing perpendicular distances

Or think: "Plug point into line equation, divide by length of normal vector $(a,b)$, take absolute value."

The denominator $\sqrt{a^2+b^2}$ is just the magnitude of the normal vector—easy to remember if you know vectors!

To find how far you need to walk, we have a magic formula. You take the equation of the path (which looks like $ax + by + c = 0$), plug in where you're standing $(x_1, y_1)$, and divide by a number that depends on the path's stepness ($\sqrt{a^2+b^2}$). The answer tells you the shortest distance!

The absolute value bars $|\cdots|$ are like saying "ignore whether the number is positive or negative—just tell me how big it is," because distance can't be negative!

#flashcards/maths

What is the formula for the distance from a point $(x_1, y_1)$ to a line $ax + by + c = 0$? :: $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$

Concept Map

Hinglish (regional understanding)