Apply d = |ax₁ + by₁ + c| / √(a² + b²).

Distance of Point from a Line ax + by + c = 0

Question

Find the distance of the point $(3, -5)$ from the line $3x - 4y + 10 = 0$ .

Solution — Step by Step

The perpendicular distance from a point $(x_1, y_1)$ to the line $ax + by + c = 0$ is:

d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}

Here $a = 3$ , $b = -4$ , $c = 10$ , and our point is $(x_1, y_1) = (3, -5)$ .

Plug into $ax_1 + by_1 + c$ :

3(3) + (-4)(-5) + 10 = 9 + 20 + 10 = 39

The absolute value gives $|39| = 39$ .

\sqrt{a^2 + b^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

A 3-4-5 Pythagorean triple — recognise it immediately to save time.

d = \frac{39}{5} = 7.8

Distance = $\dfrac{39}{5}$ units (or $7.8$ units)

Why This Works

The formula gives the perpendicular distance — the shortest possible distance from a point to a line. Any other path from the point to the line would be longer.

The denominator $\sqrt{a^2 + b^2}$ is the magnitude of the normal vector $(a, b)$ to the line. We’re essentially projecting the vector from any point on the line to $(x_1, y_1)$ onto this normal direction — that projection gives the perpendicular distance.

The absolute value in the numerator matters because $ax_1 + by_1 + c$ can be negative depending on which side of the line the point sits. Distance is always positive, so we take the magnitude.

Alternative Method — Using a Perpendicular Line

We can find the foot of the perpendicular from $(3, -5)$ to the line and compute the distance directly.

The line through $(3, -5)$ perpendicular to $3x - 4y + 10 = 0$ has slope $= -\frac{3}{-4} = \frac{3}{4}$ … wait, the original line has slope $\frac{3}{4}$ , so the perpendicular has slope $-\frac{4}{3}$ .

Perpendicular line: $y + 5 = -\frac{4}{3}(x - 3)$ , i.e., $4x + 3y + 3 = 0$ .

Solve simultaneously with $3x - 4y + 10 = 0$ :

$4x + 3y = -3$ … (i)
$3x - 4y = -10$ … (ii)

Multiply (i) by 4 and (ii) by 3: $16x + 12y = -12$ and $9x - 12y = -30$ . Adding: $25x = -42$ , so $x = -\frac{42}{25}$ .

Then $y = \frac{-3 - 4(-42/25)}{3} = \frac{-3 + 168/25}{3} = \frac{93/25}{3} = \frac{31}{25}$ .

Distance from $(3, -5)$ to $\left(-\frac{42}{25}, \frac{31}{25}\right)$ works out to $\frac{39}{5}$ — same answer.

The direct formula is always faster in exams. Use the perpendicular-foot method only when the question specifically asks for the foot of the perpendicular or the image of a point in a line.

Common Mistake

Dropping the absolute value sign. When you compute $ax_1 + by_1 + c$ , the result can be negative — for instance if the point were $(-3, 5)$ , the numerator would be $3(-3) + (-4)(5) + 10 = -9 - 20 + 10 = -19$ . A negative distance is meaningless. Always apply $|\cdot|$ before dividing. This exact slip has cost marks in CBSE board exams where method marks require correct formula application.

A second trap: misreading the sign of $b$ . The line here is $3x - 4y + 10 = 0$ , so $b = -4$ , not $+4$ . Students who rewrite it as $3x + 4y = -10$ and use $b = 4$ get $\sqrt{9 + 16} = 5$ by accident (still correct here!) but will get the numerator wrong: $3(3) + 4(-5) + 10 = 9 - 20 + 10 = -1$ , giving $d = \frac{1}{5}$ — completely wrong.

Always write out $a$ , $b$ , $c$ explicitly from the standard form $ax + by + c = 0$ before substituting.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using a Perpendicular Line

Common Mistake

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