Angle Between Lines and Distance of a Point | NIMCET

Quick answer: The acute angle θ between lines with slopes $m_1$ and $m_2$ satisfies $\tan\theta = \left\lvert\frac{m_1 - m_2}{1 + m_1 m_2}\right\rvert$ . The perpendicular distance from $(x_1, y_1)$ to $ax + by + c = 0$ is $\frac{\lvert ax_1 + by_1 + c\rvert}{\sqrt{a^2 + b^2}}$ , and the distance between parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is $\frac{\lvert c_1 - c_2\rvert}{\sqrt{a^2 + b^2}}$ . Every line through the intersection of $L_1 = 0$ and $L_2 = 0$ can be written $L_1 + \lambda L_2 = 0$ .

This continues from slope and forms of the equation of a line.

Intersection of two lines

Solve the two equations together. For $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ :

Condition	Lines are
$\frac{a_1}{a_2} \ne \frac{b_1}{b_2}$	Intersecting (one point)
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$	Parallel (no point)
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$	Coincident

Angle between two lines

$\tan\theta = \left\lvert\frac{m_1 - m_2}{1 + m_1 m_2}\right\rvert$

For lines in general form, $\tan\theta = \left\lvert\frac{a_1b_2 - a_2b_1}{a_1a_2 + b_1b_2}\right\rvert$ .

If $1 + m_1 m_2 = 0$ , the lines are perpendicular.
If $m_1 = m_2$ , they are parallel.

Distance formulas

Distance	Formula
Point $(x_1, y_1)$ to $ax + by + c = 0$	$\frac{\lvert ax_1 + by_1 + c\rvert}{\sqrt{a^2 + b^2}}$
Origin to $ax + by + c = 0$	$\frac{\lvert c\rvert}{\sqrt{a^2 + b^2}}$
Between $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$	$\frac{\lvert c_1 - c_2\rvert}{\sqrt{a^2 + b^2}}$

Before using the parallel-lines formula, make the coefficients of x and y identical in both equations.

Foot of the perpendicular and image

The foot $(h, k)$ of the perpendicular from $(x_1, y_1)$ to $ax + by + c = 0$ satisfies

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

For the image (reflection), replace the right side with $-2\frac{ax_1 + by_1 + c}{a^2 + b^2}$ .

Position of a point relative to a line

Points $(x_1, y_1)$ and $(x_2, y_2)$ are on the same side of $ax + by + c = 0$ if $ax_1 + by_1 + c$ and $ax_2 + by_2 + c$ have the same sign, and on opposite sides if the signs differ.

Concurrency and family of lines

Three lines $a_ix + b_iy + c_i = 0$ are concurrent if

$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$

(and no two are parallel). Every line through the intersection of $L_1 = 0$ and $L_2 = 0$ has the form $L_1 + \lambda L_2 = 0$ .

Worked examples

Example 1: angle

Find the acute angle between $y = 2x + 1$ and $y = -3x + 4$ .

$\tan\theta = \left\lvert\frac{2 - (-3)}{1 + 2(-3)}\right\rvert = \left\lvert\frac{5}{-5}\right\rvert = 1$

So $\theta = 45^\circ$ .

Example 2: distance of a point

Find the distance from $(3, -2)$ to $4x - 3y + 2 = 0$ .

$\frac{\lvert 12 + 6 + 2\rvert}{5} = 4$

Example 3: parallel lines

Find the distance between $3x + 4y = 5$ and $6x + 8y + 15 = 0$ .

Step 1: Rewrite the second as $3x + 4y + 7.5 = 0$ ; the first is $3x + 4y - 5 = 0$ .

Step 2: Distance $= \frac{\lvert -5 - 7.5\rvert}{5} = 2.5$ .

Example 4: line through an intersection

Find the line through the intersection of $x + y = 4$ and $x - y = 2$ that passes through $(1, 0)$ .

Step 1: $(x + y - 4) + \lambda(x - y - 2) = 0$ .

Step 2: At $(1, 0)$ : $-3 + \lambda(-1) = 0$ , so $\lambda = -3$ .

Step 3: $x + y - 4 - 3x + 3y + 6 = 0$ , so $-2x + 4y + 2 = 0$ , or $x - 2y - 1 = 0$ .

Check: the intersection is $(3, 1)$ , and $3 - 2 - 1 = 0$ ✓.

Summary table

Result	Formula
Angle	$\tan\theta = \left\lvert\frac{m_1 - m_2}{1 + m_1 m_2}\right\rvert$
Point to line	$\frac{\lvert ax_1 + by_1 + c\rvert}{\sqrt{a^2 + b^2}}$
Parallel lines	$\frac{\lvert c_1 - c_2\rvert}{\sqrt{a^2 + b^2}}$
Concurrency	Coefficient determinant $= 0$
Family through intersection	$L_1 + \lambda L_2 = 0$

Exam tips

Equalise coefficients before using the parallel-lines distance formula.
Use $L_1 + \lambda L_2 = 0$ to avoid finding the intersection point.
For images and feet of perpendiculars, the ratio formula is much faster than solving equations.

Next, read pair of straight lines and homogeneous equations, or see the NIMCET Coordinate Geometry and Vectors chapter.

Common mistakes

Dropping the absolute value and getting a negative distance.
Using the parallel-lines formula when x and y coefficients differ.
Forgetting that $\tan\theta$ gives the acute angle; the obtuse angle is $180^\circ - \theta$ .
Using the image formula for the foot of the perpendicular (it has a factor of 2).

Frequently Asked Questions

What is the formula for the angle between two lines?

tan θ = |(m₁ − m₂)/(1 + m₁m₂)|, where m₁ and m₂ are the slopes.

What is the distance of a point from a line?

The distance from (x₁, y₁) to ax + by + c = 0 is |ax₁ + by₁ + c|/√(a² + b²).

How do you find the distance between two parallel lines?

Make the x and y coefficients equal, then use |c₁ − c₂|/√(a² + b²).

When are three lines concurrent?

When the determinant of their coefficients is zero and no two of them are parallel.

What is the equation of a line through the intersection of two lines?

L₁ + λL₂ = 0, where λ is found from one more condition.