MHT-CET Maths · Straight Line

Distance — From a Point, Between Parallels, Along a Direction and the Foot of the Perpendicular

|ax₀ + by₀ + c|/√(a² + b²) is the distance from a point to a line; |c₁ − c₂|/√(a² + b²) the gap between parallels; the foot of the perpendicular is where the perpendicular through the point meets the line; distance along a direction needs the parallel line through the point.

All Straight Line notes MHT-CET Maths strategy

Why this matters

15 PYQs at 7% HARD. The stems are the nearest of four lines to the origin, points on a line at unit distance from another, the area of a square from two parallel sides (set in 2023 and 2024), the distance between two parallel lines once one is fixed by a point, the foot of the perpendicular from (−2, 3) to 3x − y = 1, a distance measured parallel to a third line, and the distance of an intersection point from the origin. One formula per stem, applied once.

Concept 1 of 4: Distance From a Point to a Line

Substitute the point into the line's expression and divide by the length of the normal vector

(a, b)

. The sign of the numerator tells which side of the line the point is on; the modulus gives the distance.

Definition

Distance of the origin from $ax + by + c = 0$ is $\dfrac{|c|}{\sqrt{a^2 + b^2}}$ : $3x - 4y + 4 = 0$ gives $0.8$ , $2x - 3y = 5$ gives $\dfrac{5}{\sqrt{13}} \approx 1.39$ , $4x - 3y + 12 = 0$ gives $2.4$ , $5x - 2y = 3$ gives $\dfrac{3}{\sqrt{29}} \approx 0.56$ — the nearest.
Points on $2x - y = 5$ at distance $1$ from $3x + 4y = 5$ : $\dfrac{|3x + 4y - 5|}{5} = 1$ with $y = 2x - 5$ : $|11x - 25| = 5$ , $x = \dfrac{30}{11}$ or $\dfrac{20}{11}$ : $\left(\dfrac{30}{11}, \dfrac{5}{11}\right)$ , $\left(\dfrac{20}{11}, -\dfrac{15}{11}\right)$ .
$x + (a - 1)y = 1$ and $2x + a^2y = 1$ perpendicular forces $a = -1$ ; they meet at $\left(\dfrac35, -\dfrac15\right)$ , at distance $\sqrt{\dfrac{2}{5}}$ from the origin — a plain two-point distance once the point is found.
Two answers to ' $|\cdot| = d$ ' are normal: one point on each side of the line.

Point to line

d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}

Worked example

Find the distance of

(3, -2)

from

5x + 12y - 4 = 0

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift I · Q105Easy

Example 1 · Straight Line · Distance — From a Point, Between Parallels, Along a Direction and the Foot of the Perpendicular

From the following options, the nearest line to the origin is ....

Comparing |c| without dividing by √(a² + b²)

5x - 2y = 3

has the smallest constant AND the smallest distance, but

3x - 4y + 4 = 0

has

|c| = 4 > 3

and distance

0.8

, less than

\dfrac{5}{\sqrt{13}}

from the line with

|c| = 5

. Always divide.

Drill 2 more on distance from a point to a line

Concept 2 of 4: Distance Between Parallel Lines, and the Square Between Them

Write both lines with the SAME

a, b

; then the gap is

\dfrac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}

. Two opposite sides of a square are parallel lines, so the side is that gap and the area its square.

Definition

$4x + 3y - 20 = 0$ and $4x + 3y + 15 = 0$ : gap $\dfrac{35}{5} = 7$ , square area $49$ . $5x - 12y + 39 = 0$ , $5x - 12y + 78 = 0$ : gap $3$ , area $9$ .
$L: \dfrac{x}{5} + \dfrac{y}{b} = 1$ through $(13, 32)$ gives $b = -20$ , i.e. $4x - y = 20$ ; $K: \dfrac{x}{c} + \dfrac{y}{3} = 1$ parallel to it is $4x - y = -3$ ; distance $\dfrac{23}{\sqrt{17}}$ .
If the coefficients differ by a factor ( $2x + y = 3$ and $4x + 2y = 1$ ), scale one line first.
The gap is also the distance from ANY point of one line to the other.

Parallel lines

d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \quad\text{for } ax + by + c_1 = 0,\ ax + by + c_2 = 0

Worked example

Two sides of a square lie on

3x + 4y = 7

and

3x + 4y = 27

. Find its area.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q135Moderate

Example 2 · Straight Line · Distance — From a Point, Between Parallels, Along a Direction and the Foot of the Perpendicular

If two sides of a square are

4x+3y-20=0

and

4x+3y+15=0

, then the area of the square is

Subtracting constants of differently scaled lines

4x + 3y = 20

and

8x + 6y = 30

are not

10

apart; rescale to

4x + 3y = 15

first, giving

1

Drill 3 more on distance between parallel lines, and the square between them

Concept 3 of 4: Foot of the Perpendicular: Intersect the Perpendicular Through the Point With the Line

The foot is on the line and on the perpendicular through the given point. Write the perpendicular (negative reciprocal slope through the point) and solve the two equations — or use the ratio form

\dfrac{h - x_1}{a} = \dfrac{k - y_1}{b} = -\dfrac{ax_1 + by_1 + c}{a^2 + b^2}

Definition

From $(-2, 3)$ to $3x - y - 1 = 0$ : the perpendicular has slope $-\tfrac13$ , $x + 3y = 7$ ; with $3x - y = 1$ : $(1, 2)$ .
Ratio form: $\dfrac{h + 2}{3} = \dfrac{k - 3}{-1} = -\dfrac{-6 - 3 - 1}{10} = 1 \Rightarrow (h, k) = (1, 2)$ .
The reflection (image) of the point is twice as far: $2 \times$ foot $-$ point $= (4, 1)$ .
A tangent's point of contact on a circle is the foot of the perpendicular from the centre — the same computation in the Circle chapter.

Foot of the perpendicular

\frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}

Worked example

Find the foot of the perpendicular from

(4, 1)

x + 2y = 6

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q140Moderate

Example 3 · Straight Line · Distance — From a Point, Between Parallels, Along a Direction and the Foot of the Perpendicular

The co-ordinates of the foot of perpendicular, drawn from the point

(-2,3)

on the line

3x-y-1=0

are

Sign slips in the ratio form

The common ratio is MINUS

\dfrac{ax_1 + by_1 + c}{a^2 + b^2}

. Dropping the minus sends the foot to the wrong side,

(-5, 4)

-style, which is never on the line — check the foot satisfies the line's equation.

Concept 4 of 4: Distance From a Point to a Line Measured Parallel to Another Line

Draw the line through the point parallel to the given direction, find where it meets the target line, and measure the ordinary distance between the point and that meeting point. It is longer than the perpendicular distance unless the direction is the normal.

Definition

From $(1, 2)$ to $x + y = 0$ parallel to $3x - y = 2$ : the parallel through $(1, 2)$ is $3x - y = 1$ ; it meets $x + y = 0$ at $\left(\tfrac14, -\tfrac14\right)$ ; distance $\sqrt{\left(\tfrac34\right)^2 + \left(\tfrac94\right)^2} = \dfrac{3\sqrt{10}}{4}$ .
Formula: if the direction has inclination $\theta$ , the distance is $\dfrac{|ax_1 + by_1 + c|}{|a\cos\theta + b\sin\theta|}$ .
The perpendicular distance is the special case $\theta =$ direction of the normal, and it is the minimum over all directions.

Along a direction

d_\theta = \frac{|ax_1 + by_1 + c|}{|a\cos\theta + b\sin\theta|}

Worked example

Find the distance of

(2, 3)

from

x - y = 5

measured parallel to the

x

-axis.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q107Moderate

Example 4 · Straight Line · Distance — From a Point, Between Parallels, Along a Direction and the Foot of the Perpendicular

The distance of the point

(1,2)

from the line

x+y= 0

measured parallel to the line

3x-y= 2

Reporting the perpendicular distance

\dfrac{3}{\sqrt2}

is the perpendicular distance from

(1, 2)

x + y = 0

; measured along

3x - y = 2

it is

\dfrac{3\sqrt{10}}{4}

. 'Measured parallel to' changes the answer.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Distance From a Point to a Line
Point to line
$d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$
Distance Between Parallel Lines, and the Square Between Them
Parallel lines
$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \quad\text{for } ax + by + c_1 = 0,\ ax + by + c_2 = 0$
Foot of the Perpendicular: Intersect the Perpendicular Through the Point With the Line
Foot of the perpendicular
$\frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$
Distance From a Point to a Line Measured Parallel to Another Line
Along a direction
$d_\theta = \frac{|ax_1 + by_1 + c|}{|a\cos\theta + b\sin\theta|}$

Watch out for (4)

Comparing |c| without dividing by √(a² + b²)
→ Distance From a Point to a Line
Subtracting constants of differently scaled lines
→ Distance Between Parallel Lines, and the Square Between Them
Sign slips in the ratio form
→ Foot of the Perpendicular: Intersect the Perpendicular Through the Point With the Line
Reporting the perpendicular distance
→ Distance From a Point to a Line Measured Parallel to Another Line

Test yourself on Straight Line

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Related notes

Circle — the tangent's point of contact is a foot of perpendicular →