MathsICSE Class 10 · Loci

Locus of Points Equidistant from Two Intersecting Lines: the Angle Bisectors (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The locus of points equidistant from two intersecting lines is the pair of lines bisecting the angles between them; the two bisectors are perpendicular. Inside one given angle, it is that angle's bisector. A point P on the bisector of an angle θ, at distance OP from the vertex, is OP sin(θ/2) from each arm.

Understand Locus Equidistant from Two Intersecting Lines

This is the second locus theorem in the CISCE syllabus: the locus of a point equidistant from two intersecting lines is the bisector of the angles between the lines (proof not required).

Why it works. If $P$ is on the bisector of $∠ X O Y$ , drop perpendiculars $P M$ and $P N$ to the arms. Triangles $O M P$ and $O N P$ have a common side $O P$ , equal angles at $O$ and right angles at $M$ and $N$ , so they are congruent (AAS) and $P M = P N$ .

Two lines make two bisectors. Two intersecting lines form two pairs of vertically opposite angles, $θ$ and $180° - θ$ . Each pair has a bisector, so the full locus is two lines. They are perpendicular, because $\frac{θ}{2} + \frac{180° - θ}{2} = 90°$ . If the question says 'inside $∠ A B C$ ' or 'equidistant from $B A$ and $B C$ ' (rays), use just the bisector of that angle.

Construction of an angle bisector.

1. With centre at the vertex $B$ , draw an arc cutting both arms at $X$ and $Y$ .
2. With centres $X$ and $Y$ and the same radius, draw arcs meeting at $Z$ .
3. Join $B Z$ . Leave the arcs visible.

The 2024 report on Q10(iii)(d) notes candidates who did not bisect $∠ A B C$ for 'equidistant from $B A$ and $B C$ '.

Distance to the arms. For $P$ on the bisector of $θ$ with $O P = l$ , the distance to each arm is $l sin \frac{θ}{2}$ .

Incentre. The bisectors of the three angles of a triangle meet at the incentre $I$ , equidistant from all three sides. The perpendicular distance from $I$ to a side is the inradius $r$ , and the circle with centre $I$ and radius $r$ (the incircle) touches all three sides. Useful checks: $r = \frac{area}{s}$ where $s$ is the semi-perimeter; for a right triangle with legs $a$ , $b$ and hypotenuse $c$ , $r = \frac{a + b - c}{2}$ .

In coordinates. Points equidistant from the x-axis and the y-axis satisfy $∣ y ∣ = ∣ x ∣$ : the lines $y = x$ and $y = - x$ .

Locus Equidistant from Two Intersecting Lines: formulas to remember

Locus theorem: $P equidistant from two intersecting lines ⟺ P lies on a bisector of the angles between them$; Proof not required.
The two bisectors: $\frac{θ}{2} + \frac{180° - θ}{2} = 90°$; They are perpendicular.
Distance from the arms: $P M = P N = O P sin \frac{θ}{2}$; $P$ on the bisector of $θ$ .
Inradius: $r = \frac{Δ}{s}; right triangle: r = \frac{a + b - c}{2}$; $Δ$ = area, $s$ = semi-perimeter.
Equidistant from the axes: $y = x or y = - x$; $∣ y ∣ = ∣ x ∣$ .

How to solve Locus Equidistant from Two Intersecting Lines questions

Decide which angle, bisect it with full arcs, then combine with any other locus.

1Identify the two lines (or rays) and the angle in which the point must lie.
2Construct the bisector of that angle (both bisectors if the lines are full lines and no region is given).
3Construct any other locus required (perpendicular bisector, circle).
4Mark the intersection and label it.
5Measure the required distances; check with $O P sin \frac{θ}{2}$ or coordinates where possible.

Locus Equidistant from Two Intersecting Lines 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.

Equidistant from two roads

Suppose two straight roads cross at an angle of $60°$ , and a bus shelter is to be placed 2 km from the crossing, the same distance from both roads, inside the $60°$ angle.

How far is the shelter from each road?

Try it first, then show the solutionSolution

It lies on the bisector of the $60°$ angle, so it makes $30°$ with each road.
Distance $= 2 sin 30° = 1$ km.

Answer: 1 km from each road.

Memory hook: On the bisector, distance to the arm = OP × sin(half-angle).

Equidistant from two walls

Imagine a lamp fixed on the floor of a rectangular room so that it is equally far from two walls that meet at a corner, and 3 m from each.

Where is it, and how far is it from the corner?

Try it first, then show the solutionSolution

The walls meet at $90°$ ; the lamp lies on the bisector, at $45°$ to each wall.
Distance from the corner $= 3^{2} + 3^{2} = 32 \approx 4.24$ m.

Answer: On the corner's bisector, about 4.24 m from the corner.

Memory hook: Two walls, one bisector.

Tricks to remember Locus Equidistant from Two Intersecting Lines

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

Two lines → bisect the angle
Two points: perpendicular bisector. Two lines: angle bisector.
Use it when: Choosing the right locus.
Full lines give an X
Two crossing lines have two bisectors, crossing at right angles like an X.
Use it when: Not missing half of the locus.
Incentre is 'in'
Angle bisectors meet at the incentre, the centre of the circle inside touching the sides.
Use it when: Telling incentre from circumcentre.

Locus Equidistant from Two Intersecting Lines: solved examples

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

Example 1: Equidistant from the axes · 1 mark · MCQBoard level

A point $P$ is equidistant from the x-axis and the y-axis. Then $P$ lies on:
(a) the x-axis
(b) $y = x$ only
(c) $y = x$ or $y = - x$
(d) a circle centred at the origin

Try it first, then show the solutionSolution

Distances are $∣ y ∣$ and $∣ x ∣$ ; $∣ y ∣ = ∣ x ∣$ gives $y = x$ or $y = - x$ , the two bisectors of the angles between the axes.

Answer: (c)

Example 2: Distance from the arms · 2 marks · Short answerBoard level

$P$ lies on the bisector of $∠ X O Y = 60°$ and is 4 cm from $O X$ . Find $O P$ and the distance of $P$ from $O Y$ .

Try it first, then show the solutionSolution

$P$ is equidistant from the arms, so it is 4 cm from $O Y$ too.
$4 = O P sin 30° = \frac{O P}{2}$ , so $O P = 8$ cm.

Answer: $O P = 8$ cm; 4 cm from $O Y$ .

Example 3: Equidistant from the axes and on a line · 3 marks · Short answerHarder board level

Find the points on the line $x + 2 y = 6$ that are equidistant from the coordinate axes.

Try it first, then show the solutionSolution

Such points lie on $y = x$ or $y = - x$ .
With $y = x$ : $3 x = 6$ , $x = 2$ : $(2, 2)$ .
With $y = - x$ : $x - 2 x = 6$ , $x = - 6$ : $(- 6, 6)$ .

Answer: $(2, 2)$ and $(- 6, 6)$

Example 4: Construction with two loci · 4 marks · Long answerHarder board level

Construct $△ A B C$ with $∠ A B C = 90°$ , $A B = 5$ cm and $B C = 6$ cm. Construct the locus of points equidistant from $B A$ and $B C$ , and the locus of points equidistant from $B$ and $C$ . Mark their intersection $P$ . Measure $B P$ and the distance of $P$ from $B C$ .

Try it first, then show the solutionSolution

Construct the right angle at $B$ and mark $A$ and $C$ .
Bisect $∠ A B C$ (each half $45°$ ), and construct the perpendicular bisector of $B C$ (3 cm from $B$ ).
They meet at $P$ , 3 cm from $B C$ and 3 cm from $B A$ .
$B P = 32 \approx 4.2$ cm.

Answer: $B P \approx 4.2$ cm; $P$ is 3 cm from $B C$ (and from $B A$ ).

Example 5: Inradius of a right triangle · 3 marks · Short answerHOTS / competency

A triangle has sides 5 cm, 12 cm and 13 cm. Find the radius of the circle that touches all three sides, using two methods.

Try it first, then show the solutionSolution

$5^{2} + 1 2^{2} = 1 3^{2}$ , so the triangle is right-angled. Area $= \frac{1}{2} \times 5 \times 12 = 30$ cm²; $s = 15$ cm.
Method 1: $r = \frac{30}{15} = 2$ cm.
Method 2: $r = \frac{5 + 12 - 13}{2} = 2$ cm.

Answer: 2 cm

Example 6: Case study: a fuel station between two roads · 4 marks · Case-basedHOTS / competency

Two straight roads cross at $O$ at an angle of $60°$ (and so also form angles of $120°$ ). A fuel station must be equally far from both roads and 2 km from $O$ .
(a) Describe the locus of points equidistant from both roads.
(b) If the station is inside a $60°$ angle, how far is it from each road?
(c) If it is inside a $120°$ angle, how far is it from each road?

Try it first, then show the solutionSolution

(a) The two bisectors of the angles between the roads; they are perpendicular.
(b) Half-angle $30°$ : $2 sin 30° = 1$ km.
(c) Half-angle $60°$ : $2 sin 60° = 3 \approx 1.73$ km.

Answer: (a) The pair of angle bisectors (b) 1 km (c) about 1.73 km

Where students lose marks in Locus Equidistant from Two Intersecting Lines

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

Mistake: Bisecting the wrong thing
'Equidistant from $B A$ and $B C$ ' needs the bisector of $∠ A B C$ , not the perpendicular bisector of a side.
Wrong working
Perpendicular bisector of $A C$ drawn.
Right working
Bisector of $∠ A B C$ .
Do this: Lines → angle bisector; points → perpendicular bisector.
Usually costs: Reported in 2024 Q10(iii)(d).
Mistake: Missing the second bisector
Two full lines form two pairs of angles.
Wrong working
Only $y = x$ for points equidistant from the axes.
Right working
$y = x$ and $y = - x$ .
Do this: Check whether the question restricts the region.
Usually costs: Half the points.
Mistake: Inaccurate base angle construction
If the triangle or angle is wrong, every later locus moves.
Wrong working
A $90°$ angle drawn by eye.
Right working
Construct $90°$ with compass arcs and keep them visible.
Do this: Construct, don't sketch.
Usually costs: Reported in 2024 Q10(iii).

Exam tips

2024 Q10(iii)(d) asked for the locus of points equidistant from $B A$ and $B C$ : the bisector of $∠ A B C$ .
Keep all construction arcs; the 2024 report asks candidates to show them.
In combined-locus questions, name each locus in a short line before drawing it.
When measuring, give lengths to the nearest millimetre.

Practice questions on Locus Equidistant from Two Intersecting Lines

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 locus of points inside $∠ P QR$ equidistant from $QP$ and $QR$ is:
(a) the perpendicular bisector of $P R$
(b) the bisector of $∠ P QR$
(c) a circle centre $Q$
(d) a line parallel to $QR$
Hint
Two arms of an angle.
Answer
(b)
Q2 · 2 marks · Short answerBoard level
$∠ X O Y = 90°$ and $P$ lies on its bisector, 6 cm from $O X$ . How far is $P$ from $O Y$ , and what is $O P$ ?
Hint
Equidistant from the arms.
Answer
6 cm; $O P = 62 \approx 8.5$ cm
Q3 · 2 marks · Short answerHarder board level
Explain why the two bisectors of the angles formed by two intersecting lines are perpendicular.
Hint
The angles are $θ$ and $180° - θ$ .
Answer
The bisectors make $\frac{θ}{2}$ and $\frac{180° - θ}{2}$ with a common line, which add up to $90°$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Every point on the bisector of $∠ A O B$ is equidistant from $O A$ and $O B$ .
Reason (R): The perpendiculars from such a point to the arms form two congruent right triangles.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
AAS with the common hypotenuse.
Answer
(c)
Q5 · 3 marks · Short answerHarder board level
In $△ A B C$ , $∠ B = 90°$ , $A B = 8$ cm and $B C = 6$ cm. The bisector of $∠ B$ meets $A C$ at $D$ . Find the distance of $D$ from $A B$ and from $B C$ .
Hint
Place $B$ at the origin; $D$ is on $y = x$ and on $\frac{x}{6} + \frac{y}{8} = 1$ .
Answer
$x (\frac{1}{6} + \frac{1}{8}) = 1$ gives $x = \frac{24}{7}$ : $D$ is $\frac{24}{7} \approx 3.43$ cm from each side.
Q6 · 4 marks · Short answerHOTS / competency
Find the points that are equidistant from the coordinate axes and also equidistant from $A (2, 0)$ and $B (0, 4)$ .
Hint
The perpendicular bisector of $A B$ is $x - 2 y + 3 = 0$ .
Answer
With $y = x$ : $(3, 3)$ . With $y = - x$ : $(- 1, 1)$ .
Q7 · 3 marks · Short answerHarder board level
Find the inradius of an equilateral triangle of side 6 cm.
Hint
Area $= \frac{3}{4} \times 36$ ; $s = 9$ .
Answer
$r = \frac{9 3}{9} = 3 \approx 1.73$ cm

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.Locus of points equidistant from two intersecting lines?Show answer
The pair of bisectors of the angles between them.
2.Angle between the two bisectors?Show answer
$90°$
3.Where do the angle bisectors of a triangle meet?Show answer
At the incentre.
4.Inradius of a right triangle with sides 6, 8, 10?Show answer
2
5.Points equidistant from the axes lie on which lines?Show answer
$y = x$ and $y = - x$

Locus Equidistant from Two Intersecting Lines: frequently asked questions

What is the locus of points equidistant from two intersecting lines?

The bisectors of the angles between the lines. There are two of them, and they are perpendicular. Inside a single given angle, use only that angle's bisector.

How do I construct an angle bisector?

From the vertex, draw an arc cutting both arms. From those two points, draw arcs of equal radius that meet. Join the vertex to the meeting point.

What is the incentre?

The point where the three angle bisectors of a triangle meet. It is equidistant from the three sides and is the centre of the incircle.