MathsISC Class 11 · Straight Lines

General Equation, Distances, Angle Bisectors and the Locus of a Point

Updated 17 October 2026 · Fact-checked

The distance of (x₁, y₁) from Ax + By + C = 0 is |Ax₁ + By₁ + C|/√(A² + B²); parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 are |C₁ − C₂|/√(A² + B²) apart. Parallel lines share the coefficients A and B; perpendicular lines have Bx − Ay + k = 0. A locus is the path of a point satisfying a given condition, found by translating the condition into an equation in x and y.

The shortest distance d is the perpendicular from P to the line; all points at that same distance lie on two parallel lines.

Illustration drawn for QuestionBuddy.

Understand Distance from a Line, Angle Bisectors and Locus

General equation. $A x + B y + C = 0$ has slope $- A / B$ , x-intercept $- C / A$ and y-intercept $- C / B$ . A line parallel to it has the same A and B and a different C; a line perpendicular to it has the form $B x - A y + k = 0$ (swap the coefficients and change one sign).

Distance of a point from a line. $d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$ . The sign inside the modulus tells which side of the line the point lies on. For parallel lines $A x + B y + C_{1} = 0$ and $A x + B y + C_{2} = 0$ (same A and B): $d = \frac{∣ C _{1} - C _{2} ∣}{A ^{2} + B ^{2}}$ .

Angle bisectors. The points equidistant from $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ satisfy $\frac{a _{1} x + b _{1} y + c _{1}}{a _{1}^{2} + b _{1}^{2}} = \pm \frac{a _{2} x + b _{2} y + c _{2}}{a _{2}^{2} + b _{2}^{2}}$ , which gives two perpendicular bisectors.

Families of lines. The line $L_{1} + λ L_{2} = 0$ passes through the intersection of $L_{1} = 0$ and $L_{2} = 0$ for every value of λ. Find λ from one more condition.

Locus. (1) Take a general point P(x, y). (2) Write the given condition using the distance formula, slope or area. (3) Simplify to a relation in x and y. Examples: the locus of a point equidistant from two points is the perpendicular bisector of the segment joining them; the locus of a point at distance k from a line is a pair of parallel lines.

Distance from a Line, Angle Bisectors and Locus: formulas to remember

Point to line: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$; Absolute value gives a positive distance.
Parallel lines: $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$; Same A and B.
Perpendicular line: $Given $Ax + By + C = 0$, take $Bx - Ay + k = 0$$; Slopes multiply to −1.
Bisectors: $\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = ±\frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$; Two bisectors, one for each sign.

Distance from a Line, Angle Bisectors and Locus in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Distance of a house from a straight road

Suppose a straight road follows the line $3 x + 4 y - 5 = 0$ (units in kilometres) and a house is at (1, 2).

Find the shortest distance from the house to the road.

Try it first, then show the solutionSolution

Use $d = \frac{∣3 ( 1 ) + 4 ( 2 ) - 5∣}{3 ^{2} + 4 ^{2}}$ .
Numerator: $∣3 + 8 - 5∣ = 6$ ; denominator: $25 = 5$ .
$d = \frac{6}{5} = 1.2$ km.

Answer: 1.2 km.

Memory hook: The perpendicular distance is the shortest.

Tricks to remember Distance from a Line, Angle Bisectors and Locus

Short memory aids. Say each one aloud once, then use it on a question the same day.

Swap and negate
The perpendicular of Ax + By + C = 0 is Bx − Ay + k = 0.
Use it when: Perpendicular lines through a point.
Parallel: change only C
Parallel lines have the same A and B.
Use it when: Parallel lines.

Distance from a Line, Angle Bisectors and Locus: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Distance from a point · 1 mark · MCQBoard level

The distance of (1, 2) from the line $3 x + 4 y - 5 = 0$ is:
(a) $\frac{6}{5}$
(b) 6
(c) $\frac{5}{6}$
(d) 1

Try it first, then show the solutionSolution

$\frac{∣3 + 8 - 5∣}{5} = \frac{6}{5}$ .

Answer: (a) $\frac{6}{5}$

Example 2: Parallel and perpendicular lines · 3 marks · Short answerBoard level

Find the equations of the lines through (2, 3) parallel and perpendicular to $3 x - 2 y + 5 = 0$ .

Try it first, then show the solutionSolution

Parallel: $3 x - 2 y + c = 0$ through (2, 3): $6 - 6 + c = 0$ , so $c = 0$ : $3 x - 2 y = 0$ . (1.5 marks)
Perpendicular: $2 x + 3 y + k = 0$ through (2, 3): $4 + 9 + k = 0$ , so $k = - 13$ . (1.5 marks)

Answer: 3x − 2y = 0; 2x + 3y − 13 = 0.

Example 3: Distance between parallel lines · 3 marks · Short answerHarder board level

Find the distance between the lines $3 x + 4 y + 5 = 0$ and $3 x + 4 y - 10 = 0$ , and the equation of the line midway between them.

Try it first, then show the solutionSolution

$d = \frac{∣5 - ( - 10 ) ∣}{9 + 16} = \frac{15}{5} = 3$ . (1.5 marks)
The midway line has $C = \frac{5 + ( - 10 )}{2} = - 2.5$ : $3 x + 4 y - 2.5 = 0$ or $6 x + 8 y - 5 = 0$ . (1.5 marks)

Answer: Distance 3; midway line 6x + 8y − 5 = 0.

Example 4: Angle bisectors · 4 marks · Short answerHarder board level

Find the equations of the bisectors of the angles between the lines $3 x - 4 y + 7 = 0$ and $5 x + 12 y - 3 = 0$ .

Try it first, then show the solutionSolution

$\frac{3 x - 4 y + 7}{5} = \pm \frac{5 x + 12 y - 3}{13}$ . (1 mark)
Plus sign: $13 (3 x - 4 y + 7) = 5 (5 x + 12 y - 3)$ , i.e. $14 x - 112 y + 106 = 0$ , or $7 x - 56 y + 53 = 0$ . (1.5 marks)
Minus sign: $39 x - 52 y + 91 = - 25 x - 60 y + 15$ , i.e. $64 x + 8 y + 76 = 0$ , or $16 x + 2 y + 19 = 0$ . (1.5 marks)

Answer: 7x − 56y + 53 = 0 and 16x + 2y + 19 = 0.

Example 5: Locus problems · 5 marks · Long answerHOTS / competency

Find the locus of a point P(x, y) which is equidistant from A(1, 2) and B(5, −2). Find also the locus of points at distance 2 from the line $3 x + 4 y = 5$ . What is the angle between your first locus and the line AB?

Try it first, then show the solutionSolution

$P A^{2} = P B^{2}$ : $(x - 1)^{2} + (y - 2)^{2} = (x - 5)^{2} + (y + 2)^{2}$ . (1 mark)
Expand: $- 2 x - 4 y + 5 = - 10 x + 4 y + 29$ , so $8 x - 8 y = 24$ , i.e. $x - y = 3$ . (1.5 marks)
Distance 2 from $3 x + 4 y - 5 = 0$ : $∣3 x + 4 y - 5∣ = 10$ , so $3 x + 4 y - 15 = 0$ or $3 x + 4 y + 5 = 0$ . (1.5 marks)
Slope of AB $= \frac{- 2 - 2}{5 - 1} = - 1$ ; the locus has slope 1; $(- 1) (1) = - 1$ , so they are perpendicular (90°). (1 mark)

Answer: x − y = 3; 3x + 4y − 15 = 0 and 3x + 4y + 5 = 0; the angle is 90°.

Where students lose marks in Distance from a Line, Angle Bisectors and Locus

The slips examiners see most often, and what to write instead.

Mistake: Missing the modulus in the distance formula
Giving a negative distance.
Do this: Distance is the absolute value.
Usually costs: 1 mark.
Mistake: Unequal A and B before using the parallel formula
Applying it to 3x + 4y and 6x + 8y without scaling.
Do this: Make A and B equal first.
Usually costs: 2 marks.
Mistake: Wrong perpendicular line
Using Ax − By + k.
Do this: The perpendicular is Bx − Ay + k = 0.
Usually costs: 1 mark.

Exam tips

Put equations in the form Ax + By + C = 0.
Scale parallel lines to the same A and B.
Check distances by an alternative method.
For locus problems, simplify to a recognisable curve.

Practice questions on Distance from a Line, Angle Bisectors and Locus

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The line perpendicular to $2 x + 3 y = 6$ through the origin is:
(a) $2 x + 3 y = 0$
(b) $3 x - 2 y = 0$
(c) $3 x + 2 y = 0$
(d) $2 x - 3 y = 0$
Hint
Swap the coefficients and change one sign.
Answer
(b) $3 x - 2 y = 0$
Q2 · 2 marks · Short answerBoard level
Find the distance of the origin from $5 x - 12 y + 26 = 0$ .
Hint
Substitute (0, 0).
Answer
$\frac{26}{13} = 2$ .
Q3 · 3 marks · Short answerHarder board level
Find the distance of the point (3, −2) from the line $4 x - 3 y + 5 = 0$ and say on which side of the line it lies compared with the origin.
Hint
Compare signs.
Answer
Value at (3, −2): $12 + 6 + 5 = 23$ , so $d = \frac{23}{5} = 4.6$ . At the origin the value is $+ 5$ , also positive, so both are on the same side.
Q4 · 3 marks · Short answerHarder board level
Find the equation of the line through the intersection of $x + y = 4$ and $2 x - y = 2$ and the point (1, 0).
Hint
Use L₁ + λL₂ = 0.
Answer
Intersection: $x = 2$ , $y = 2$ . Slope through (2, 2) and (1, 0) is 2: $y = 2 (x - 1)$ , i.e. $2 x - y - 2 = 0$ .
Q5 · 3 marks · Short answerHarder board level
Find the locus of a point whose distance from the point (2, 0) is equal to its distance from the line $x = - 2$ .
Hint
Equate the two distances.
Answer
$(x - 2)^{2} + y^{2} = (x + 2)^{2}$ , so $y^{2} = 8 x$ (a parabola).
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a delivery drone must fly to a point on the line $x - y = 4$ nearest to a warehouse at (0, 0) in kilometres.
(i) Find the shortest distance. (1 mark)
(ii) Find the equation of the perpendicular from the warehouse. (1 mark)
(iii) Find the foot of the perpendicular. (1 mark)
(iv) Check the distance using the foot. (1 mark)
Hint
The perpendicular to x − y = 4 has the form x + y = k.
Answer
(i) $d = \frac{∣0 - 0 - 4∣}{2} = 22 \approx 2.83$ km. (ii) $x + y = 0$ . (iii) Solving: $x = 2$ , $y = - 2$ : foot (2, −2). (iv) $4 + 4 = 22$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Distance formula?Show answer
|Ax₁ + By₁ + C|/√(A² + B²).
2.Perpendicular of Ax + By + C = 0?Show answer
Bx − Ay + k = 0.
3.Locus equidistant from two points?Show answer
The perpendicular bisector.
4.Number of angle bisectors?Show answer
Two.

Distance from a Line, Angle Bisectors and Locus: frequently asked questions

How do I know which bisector is which?

The bisector containing the origin side (or the acute angle) is found by checking the sign of c₁c₂ and the quantity a₁a₂ + b₁b₂.

What is a locus?

The set of all points that satisfy a given geometric condition.

Why is the shortest distance perpendicular?

By Pythagoras, any other path from the point to the line is the hypotenuse of a right triangle and so longer.