Distance of a Point from a Line

The Perpendicular Distance Formula

The distance of a point from a line always means the perpendicular distance — the length of the shortest segment from the point to the line.

Key Point (Distance of a point from a line): The distance of $P(x_1, y_1)$ from the line $Ax + By + C = 0$ is

$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

The derivation (triangle-area argument). Let the line meet the axes at $Q\left(-\frac{C}{A}, 0\right)$ and $R\left(0, -\frac{C}{B}\right)$. The perpendicular $PM$ from $P$ to the line is the height of triangle $PQR$ on base $QR$, so

$$\text{area}(\triangle PQR) = \frac{1}{2}\, QR \cdot PM \quad\Rightarrow\quad PM = \frac{2\,\text{area}(\triangle PQR)}{QR}$$

Computing the area from the three coordinates gives $\frac{|C|}{2|AB|}\,|Ax_1 + By_1 + C|$, and $QR = \frac{|C|}{|AB|}\sqrt{A^2 + B^2}$. Dividing, everything except the formula cancels:

$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

How to use it: put the line in general form first, substitute the point into the left side, take the absolute value, divide by $\sqrt{A^2 + B^2}$.

For example: the distance of $(3, -5)$ from $3x - 4y - 26 = 0$:

$$d = \frac{|3(3) - 4(-5) - 26|}{\sqrt{3^2 + (-4)^2}} = \frac{|9 + 20 - 26|}{5} = \frac{3}{5}$$

Special case: the distance of the origin from $Ax + By + C = 0$ is simply $\frac{|C|}{\sqrt{A^2 + B^2}}$.

The absolute value is essential: $Ax_1 + By_1 + C$ is positive for points on one side of the line and negative on the other, but a distance must come out non-negative either way.

Distance Between Parallel Lines

Parallel lines can be written with the same $A$ and $B$: $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$. Picking any point on one line and dropping a perpendicular to the other gives:

Key Point (Distance between parallel lines):

$$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$$

For example: the lines $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$ are at distance

$$d = \frac{|7 - 5|}{\sqrt{9 + 16}} = \frac{2}{5}$$

Warning: the formula needs matching coefficients. For $3x - 4y + 7 = 0$ and $6x - 8y + 5 = 0$, first divide the second by 2 to get $3x - 4y + \frac{5}{2} = 0$, and only then apply the formula: $d = \frac{|7 - \frac{5}{2}|}{5} = \frac{9}{10}$.

Two constructions used constantly:

A line parallel to $Ax + By + C = 0$ through $(x_1, y_1)$ keeps the same $A, B$:

$$A(x - x_1) + B(y - y_1) = 0$$

A line perpendicular to $Ax + By + C = 0$ through $(x_1, y_1)$ swaps the coefficients and one sign:

$$B(x - x_1) - A(y - y_1) = 0$$

Both follow from slopes: the given line has slope $-\frac{A}{B}$, so a parallel line repeats it and a perpendicular line uses $\frac{B}{A}$.

[Board Tip] In distance calculations, the single most common slip is forgetting to move the equation into general form first. $12(x + 6) = 5(y - 2)$ must become $12x - 5y + 82 = 0$ before substituting the point — substituting into a non-standard form gives garbage that still looks like a number.

Solved Examples

Example 1: Reduce to slope-intercept form

Reduce to slope-intercept form and read off slope and $y$-intercept: (i) $x + 7y = 0$ (ii) $6x + 3y - 5 = 0$ (iii) $y = 0$.

Solution:

Step 1 — (i) Solve for $y$. $y = -\frac{1}{7}x$: slope $-\frac{1}{7}$, $y$-intercept $0$.

Step 2 — (ii) Solve for $y$. $y = -2x + \frac{5}{3}$: slope $-2$, $y$-intercept $\frac{5}{3}$.

Step 3 — (iii) Read directly. $y = 0 \cdot x + 0$: slope 0, intercept 0 — the $x$-axis itself.

Takeaway: "Reduce" just means solve for $y$ — the two numbers you need then sit in plain sight.

Example 2: Reduce to intercept form

Reduce to intercept form and find the intercepts: (i) $3x + 2y - 12 = 0$ (ii) $4x - 3y = 6$ (iii) $3y + 2 = 0$.

Solution:

Step 1 — (i) Divide by 12. $\frac{x}{4} + \frac{y}{6} = 1$: intercepts 4 and 6.

Step 2 — (ii) Divide by 6. $\frac{x}{3/2} + \frac{y}{-2} = 1$: intercepts $\frac{3}{2}$ and $-2$.

Step 3 — (iii) Recognise the special case. $y = -\frac{2}{3}$ is parallel to the $x$-axis — no $x$-intercept, hence NO intercept form.

Takeaway: Divide by the constant to force a 1 on the right; horizontal and vertical lines refuse the form entirely.

Example 3: First distance computation

Find the distance of the point $(3, -5)$ from the line $3x - 4y - 26 = 0$.

Solution:

Step 1 — Substitute into the numerator. $3(3) - 4(-5) - 26 = 9 + 20 - 26 = 3$.

Step 2 — Compute the denominator. $\sqrt{3^2 + (-4)^2} = 5$.

Step 3 — Divide. $d = \frac{|3|}{5} = \frac{3}{5}$.

Takeaway: Numerator: plug the point in; denominator: root of the coefficient squares — nothing else.

Example 4: Put it in general form first

Find the distance of the point $(-1, 1)$ from the line $12(x + 6) = 5(y - 2)$.

Solution:

Step 1 — Convert to general form. $12x + 72 = 5y - 10$, i.e. $12x - 5y + 82 = 0$.

Step 2 — Substitute. $12(-1) - 5(1) + 82 = -12 - 5 + 82 = 65$.

Step 3 — Divide. $d = \frac{65}{\sqrt{144 + 25}} = \frac{65}{13} = 5$.

Takeaway: ALWAYS move to $Ax + By + C = 0$ before substituting — a non-standard form gives a wrong number that looks right.

Example 5: Points at a given distance

Find the points on the $x$-axis whose distances from the line $\frac{x}{3} + \frac{y}{4} = 1$ are 4 units.

Solution:

Step 1 — General form. Multiply by 12: $4x + 3y - 12 = 0$.

Step 2 — Set up the distance equation. For $(a, 0)$: $\frac{|4a - 12|}{5} = 4$.

Step 3 — Solve both signs. $4a - 12 = \pm 20$: $a = 8$ or $a = -2$.

Step 4 — Report both points. $(8, 0)$ and $(-2, 0)$ — one on each side of the line.

Takeaway: Removing an absolute value always spawns two cases — geometrically, one point on each side.

Example 6: Parallel lines

Find the distance between (i) $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$, (ii) $15x + 8y - 34 = 0$ and $15x + 8y + 31 = 0$, (iii) $l(x + y) + p = 0$ and $l(x + y) - r = 0$.

Solution:

Step 1 — (i). Matching coefficients already: $d = \frac{|7 - 5|}{\sqrt{9 + 16}} = \frac{2}{5}$.

Step 2 — (ii). $\sqrt{225 + 64} = 17$: $d = \frac{|-34 - 31|}{17} = \frac{65}{17}$.

Step 3 — (iii). Coefficients $l, l$: $d = \frac{|p - (-r)|}{\sqrt{2l^2}} = \frac{|p + r|}{\sqrt{2}\,|l|}$.

Takeaway: The formula is one subtraction once the coefficient pairs match — check that first, every time.

Example 7: Parallel and perpendicular constructions

(i) Find the line parallel to $3x - 4y + 2 = 0$ through $(-2, 3)$. (ii) Find the line perpendicular to $x - 7y + 5 = 0$ with $x$-intercept 3.

Solution:

Step 1 — (i) Keep the coefficients. $3(x + 2) - 4(y - 3) = 0$, i.e. $3x - 4y + 18 = 0$.

Step 2 — (ii) Perpendicular slope. Given slope $\frac{1}{7}$ → perpendicular slope $-7$, through $(3, 0)$.

Step 3 — (ii) Write it. $y = -7(x - 3)$, i.e. $7x + y - 21 = 0$.

Takeaway: Parallel: same $A, B$, adjust $C$ via the point. Perpendicular: swap coefficients, flip one sign.

Example 8: Angle and a right-angle condition

(i) Find the angles between $\sqrt{3}x + y = 1$ and $x + \sqrt{3}y = 1$. (ii) The line through $(h, 3)$ and $(4, 1)$ meets $7x - 9y - 19 = 0$ at right angles: find $h$.

Solution:

Step 1 — (i) Slopes. $-\sqrt{3}$ and $-\frac{1}{\sqrt{3}}$.

Step 2 — (i) Angle formula. $\tan\theta = \left|\frac{-\frac{1}{\sqrt{3}} + \sqrt{3}}{1 + 1}\right| = \frac{2/\sqrt{3}}{2} = \frac{1}{\sqrt{3}}$: $\theta = 30°$, and the obtuse partner is $150°$.

Step 3 — (ii) Perpendicularity. $\frac{1 - 3}{4 - h} \cdot \frac{7}{9} = -1$: $\frac{-14}{9(4 - h)} = -1$ gives $9(4 - h) = 14$, so $h = \frac{22}{9}$.

Takeaway: Report both the acute and obtuse angles unless the question specifies — they always pair to $180°$.

Example 9: The parallel-line identity

Prove that the line through $(x_1, y_1)$ parallel to $Ax + By + C = 0$ is $A(x - x_1) + B(y - y_1) = 0$.

Solution:

Step 1 — Same slope. The proposed equation has coefficients $A, B$, hence slope $-\frac{A}{B}$ — parallel to the given line.

Step 2 — Passes through the point. Substituting $(x_1, y_1)$: $A \cdot 0 + B \cdot 0 = 0$ ✓.

Step 3 — Conclude. A line with the right slope through the right point — exactly the claim. ∎

Takeaway: This identity is the fastest parallel-line constructor — no slope computation, no rearrangement.

Example 10: Right bisector

Find the equation of the right bisector of the segment joining $(3, 4)$ and $(-1, 2)$.

Solution:

Step 1 — Midpoint. $(1, 3)$.

Step 2 — Perpendicular slope. Segment slope $\frac{2-4}{-1-3} = \frac{1}{2}$; bisector slope $-2$.

Step 3 — Point-slope. $y - 3 = -2(x - 1)$, i.e. $2x + y - 5 = 0$.

Takeaway: Right bisector = midpoint + negative-reciprocal slope — two ingredients, one line.

Example 11: Foot of the perpendicular

Find the foot of the perpendicular from $(-1, 3)$ to the line $3x - 4y - 16 = 0$.

Solution:

Step 1 — Write the perpendicular line. Given slope $\frac{3}{4}$ → perpendicular slope $-\frac{4}{3}$ through $(-1, 3)$: $y - 3 = -\frac{4}{3}(x + 1)$, i.e. $4x + 3y - 5 = 0$.

Step 2 — Intersect the two lines. Solve $3x - 4y = 16$ and $4x + 3y = 5$: multiply and add to eliminate — $x = \frac{68}{25}$, $y = -\frac{49}{25}$.

Step 3 — Conclude. The foot is $\left(\frac{68}{25}, -\frac{49}{25}\right)$.

Takeaway: Foot of perpendicular = intersection of the line with its perpendicular through the point.

Example 12: Perpendicular data

(i) The perpendicular from the origin to $y = mx + c$ meets it at $(-1, 2)$: find $m$ and $c$. (ii) In $\triangle ABC$ with $A(2, 3)$, $B(4, -1)$, $C(1, 2)$, find the equation and length of the altitude from $A$.

Solution:

Step 1 — (i) Slope from the foot. Origin to $(-1, 2)$: slope $-2$; the line is perpendicular: $m = \frac{1}{2}$.

Step 2 — (i) Find $c$. $(-1, 2)$ on the line: $2 = -\frac{1}{2} + c$, so $c = \frac{5}{2}$.

Step 3 — (ii) Altitude's equation. $BC$ has slope $\frac{2+1}{1-4} = -1$; altitude slope 1 through $A$: $x - y + 1 = 0$.

Step 4 — (ii) Altitude's length. Line $BC$: $x + y - 3 = 0$; distance from $A(2,3)$: $\frac{|2 + 3 - 3|}{\sqrt{2}} = \sqrt{2}$.

Takeaway: An altitude is a perpendicular construction plus a point-to-line distance — both from this section's toolkit.

Example 13: A clean distance identity

If $p$ is the distance of the origin from the line with intercepts $a$ and $b$, show that $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.

Solution:

Step 1 — General form of the intercept line. $\frac{x}{a} + \frac{y}{b} = 1$ becomes $bx + ay - ab = 0$.

Step 2 — Distance from the origin. $p = \frac{|{-ab}|}{\sqrt{a^2 + b^2}} = \frac{|ab|}{\sqrt{a^2 + b^2}}$.

Step 3 — Square and invert. $\frac{1}{p^2} = \frac{a^2 + b^2}{a^2b^2} = \frac{1}{b^2} + \frac{1}{a^2}$. ∎

Takeaway: Origin distances need only $\frac{|C|}{\sqrt{A^2+B^2}}$ — the identity is one clean algebraic split.