Chapter 7 of 14

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Rajasthan (RBSE)Class 10 Mathematics★ 4–6 marks · RBSE Class 10 Maths board (Unit — Coordinate Geometry)

Coordinate Geometry

RBSE Class 10 Maths on the geometry of points — the distance formula, the section formula (and midpoint), and finding the area of a triangle from its vertices. Learn to prove figures, check collinearity and divide a segment in a given ratio, with fully solved board-pattern questions.

⏱ 28 min readEasy difficulty✓ Free · NCERT-aligned

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Every concept and exam-favourite branch — revisable in 5 minutes.

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Study Guide

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Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1Find the distance between two points and from the origin
2Use the section formula and midpoint formula, and find the dividing ratio
3Compute the area of a triangle from its vertices and test collinearity
4Classify triangles and quadrilaterals using side and diagonal lengths
5Find unknown coordinates from distance, midpoint or ratio conditions

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Why this chapter matters

A formula-driven, high-certainty chapter. The distance and section formulas appear almost every year, and area/collinearity gives quick, fully scorable questions with straightforward arithmetic.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Coordinate Geometry (Class 9)

The Cartesian plane, quadrants and plotting points.

Coordinate Geometry — RBSE Class 10 (Mathematics)

Give every point an address — an (x, y) pair — and geometry becomes arithmetic. Distances, midpoints, and even the area of a triangle can now be computed from coordinates instead of measured with a ruler. This short, formula-driven chapter is among the easiest places to score full marks.

1. The distance formula

The distance between $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ is $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$

It is just the Pythagoras theorem applied to the horizontal and vertical gaps. Distance from the origin to $(x, y)$ is $x^{2} + y^{2}$ .

Uses: show a triangle is isosceles/equilateral (compare side lengths), a quadrilateral is a parallelogram/rhombus/square (compare sides and diagonals), or three points are collinear (the two shorter distances sum to the longest).

2. The section formula

The point that divides the join of $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ internally in the ratio $m_{1} : m_{2}$ is $P = (\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}})$

Midpoint (ratio 1 : 1)

$M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$

To find the ratio in which a point (or an axis) divides a segment, set $m_{1} : m_{2} = k : 1$ and solve using one coordinate.

3. Area of a triangle

For vertices $A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})$ : $Area = \frac{1}{2} x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})$

The absolute value keeps the area positive. A powerful corollary: $three points are collinear ⟺ this area = 0.$

4. Exam-favourite tasks

Classify a quadrilateral: compute all four sides and both diagonals — equal sides + equal diagonals ⇒ square; equal sides, unequal diagonals ⇒ rhombus; etc.
Find an unknown coordinate given a distance or that a point is the midpoint.
Point on an axis: a point on the x-axis is $(x, 0)$ ; on the y-axis is $(0, y)$ — use this to cut the unknowns to one.
Centroid of a triangle: $(\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})$ .

5. Worked idea

Show that $A (1, 7), B (4, 2), C (- 1, - 1), D (- 4, 4)$ form a square. Each side works out to $34$ (e.g. $A B = 9 + 25 = 34$ ), and both diagonals equal $68$ . Equal sides and equal diagonals ⇒ square.

6. Closing thought

Three formulas — distance, section, area — cover the whole chapter, and each is a direct substitution. The skill is choosing the right one and, for proofs, saying why the numbers prove the shape. In the RBSE board this is high-certainty territory: expect a distance/section question and often an area-or-collinearity problem, all fully scorable with careful arithmetic.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Distance formula

PQ = √[(x₂−x₁)² + (y₂−y₁)²]

Pythagoras on the coordinate gaps.

Distance from origin

√(x² + y²)

Special case with (0,0).

Section formula (internal)

((m₁x₂+m₂x₁)/(m₁+m₂), (m₁y₂+m₂y₁)/(m₁+m₂))

Divides AB in ratio m₁:m₂.

Midpoint

((x₁+x₂)/2, (y₁+y₂)/2)

Section formula with 1:1.

Area of triangle

½|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)|

Area 0 ⇒ collinear.

Centroid

((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)

Mean of the vertices.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Dropping the square root or the square in the distance formula

✓ Square each gap, add, THEN take the root. (x₂−x₁)² not (x₂−x₁).

WATCH OUT

✗ Swapping the terms in the section formula

✓ The x₂ pairs with m₁ and x₁ with m₂: (m₁x₂ + m₂x₁)/(m₁+m₂). Getting it reversed gives the wrong point.

WATCH OUT

✗ Forgetting the modulus in the area formula

✓ Area is always positive — wrap the expression in |…|. A negative value just means the vertices were taken clockwise.

WATCH OUT

✗ Concluding a shape without checking diagonals

✓ Equal sides alone give a rhombus; you must also compare diagonals to distinguish a square from a rhombus.

WATCH OUT

✗ Not using (x,0) or (0,y) for axis points

✓ A point on the x-axis has y = 0; on the y-axis x = 0 — this removes one unknown immediately.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 7.1

Exercise 7.1

Distance formula; collinearity; type of triangle/quadrilateral

Questions

Ex 7.2

Exercise 7.2

Section formula, midpoint and dividing ratios

Questions

Examples (in-text)

Worked distance, section and area problems

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Distance

Find the distance between (2, 3) and (5, 7).

▶ Show solution

Step 1 — √[(5−2)² + (7−3)²] = √(9 + 16) = √25.

✦ Answer: 5 units.

Q2EASY· Midpoint

Find the midpoint of the segment joining (−2, 4) and (6, −8).

▶ Show solution

Step 1 — ((−2+6)/2, (4−8)/2) = (2, −2).

✦ Answer: (2, −2).

Q3EASY· Collinearity

Are (1, 1), (2, 3) and (3, 5) collinear?

▶ Show solution

Step 1 — Area = ½|1(3−5) + 2(5−1) + 3(1−3)| = ½|−2 + 8 − 6| = 0.

✦ Answer: Yes, collinear (area = 0).

Q4MEDIUM· Section formula

Find the point dividing the join of (−1, 7) and (4, −3) in the ratio 2 : 3.

▶ Show solution

Step 1 — x = (2·4 + 3·(−1))/5 = (8−3)/5 = 1.

Step 2 — y = (2·(−3) + 3·7)/5 = (−6+21)/5 = 3.

✦ Answer: (1, 3).

Q5MEDIUM· Area

Find the area of the triangle with vertices (0, 0), (4, 0) and (0, 3).

▶ Show solution

Step 1 — ½|0(0−3) + 4(3−0) + 0(0−0)| = ½|12| = 6.

✦ Answer: 6 square units.

Q6MEDIUM· Unknown coordinate

Find x if the distance between (x, 2) and (3, 6) is 5 units.

▶ Show solution

Step 1 — (x−3)² + (6−2)² = 25 ⇒ (x−3)² + 16 = 25.

Step 2 — (x−3)² = 9 ⇒ x − 3 = ±3 ⇒ x = 6 or x = 0.

✦ Answer: x = 6 or x = 0.

Q7MEDIUM· Dividing ratio

In what ratio does the x-axis divide the segment joining (2, −3) and (5, 6)?

▶ Show solution

Step 1 — On the x-axis, y = 0. Let ratio k:1. y = (6k + (−3))/(k+1) = 0.

Step 2 — 6k − 3 = 0 ⇒ k = 1/2, so ratio = 1 : 2.

✦ Answer: 1 : 2.

Q8HARD· Shape proof

Show that A(1, 7), B(4, 2), C(−1, −1), D(−4, 4) are the vertices of a square.

▶ Show solution

Step 1 — AB = √(9+25) = √34; BC = √(25+9) = √34; CD = √(9+25) = √34; DA = √(25+9) = √34 (all sides equal).

Step 2 — Diagonals: AC = √(4+64) = √68; BD = √(64+4) = √68 (equal).

Step 3 — Equal sides AND equal diagonals ⇒ square.

✦ Answer: it is a square (side √34, diagonal √68).

Q9HARD· Find vertex

Two vertices of a triangle are (3, −5) and (−7, 4). If its centroid is (2, −1), find the third vertex.

▶ Show solution

Step 1 — Centroid x: (3 + (−7) + x₃)/3 = 2 ⇒ −4 + x₃ = 6 ⇒ x₃ = 10.

Step 2 — Centroid y: (−5 + 4 + y₃)/3 = −1 ⇒ −1 + y₃ = −3 ⇒ y₃ = −2.

✦ Answer: (10, −2).

Q10HARD· Area from k

Find k if the points (7, −2), (5, 1) and (3, k) are collinear.

▶ Show solution

Step 1 — Area = 0: 7(1−k) + 5(k−(−2)) + 3(−2−1) = 0.

Step 2 — 7 − 7k + 5k + 10 − 9 = 0 ⇒ −2k + 8 = 0 ⇒ k = 4.

✦ Answer: k = 4.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Distance: √[(x₂−x₁)² + (y₂−y₁)²].
•Section formula divides AB in m₁:m₂; midpoint is the 1:1 case.
•Area = ½|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)|.
•Area = 0 ⇔ points collinear.
•Centroid = mean of the three vertices.
•Square: equal sides and equal diagonals; rhombus: equal sides, unequal diagonals.
•Point on x-axis is (x,0); on y-axis is (0,y).

Rajasthan (RBSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 4–6 marks

Question type	Marks each	Typical count	What it tests
MCQ / very short	1	1	Distance, midpoint, collinearity
Short answer	2–3	1	Section formula or area of triangle
Long answer	3–4	0–1	Shape proof or find-a-vertex

Prep strategy

Memorise distance, section and area formulas cold
For shape proofs, always compute BOTH sides and diagonals
Use (x,0)/(0,y) to reduce unknowns for axis points
Keep the modulus in the area formula and reject negatives

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

GPS and navigation

Distances between map coordinates use the distance formula.

Computer graphics

Midpoints and section points position and interpolate pixels and vertices.

Surveying and land area

The vertex-area formula computes plot areas from corner coordinates.

Robotics and CAD

Path planning divides segments in set ratios to place waypoints.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Write the formula, substitute, then simplify — on separate lines for method marks.
For shape proofs, tabulate the four sides and two diagonals before concluding.
Always keep the modulus in area and reject a negative result.
Use axis conditions (y=0 or x=0) to cut unknowns.
State the final shape/answer explicitly with a one-line reason.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

The equation of a straight line and slope from two points.
Distance of a point from a line and the foot of the perpendicular.
Section formula for external division and harmonic conjugates.
Area of any polygon by the shoelace (surveyor's) formula.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

RBSE Class 10 Board (BSER Ajmer)High — distance/section question almost every year

NTSE / state scholarshipMedium — coordinate MCQs

JEE FoundationHigh — coordinate geometry expands greatly in Class 11

Maths Olympiad (IMO/NMTC)Medium — coordinate methods for geometry

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Yes — RBSE (BSER, Ajmer) prescribes the NCERT Mathematics textbook; the chapters match the national syllabus while RBSE sets its own exam pattern.

Assume the ratio is k:1, put it into the section formula for the known coordinate, and solve for k. Convert k to k:1.

Compute the triangle area with the three points. If it equals zero, they lie on one straight line.

Both have all four sides equal. A square also has equal diagonals; a rhombus has unequal diagonals. Always check the diagonals.

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