What is the distance formula in coordinate geometry?

The distance between two points P(x₁,y₁) and Q(x₂,y₂) is PQ = √[(x₂−x₁)² + (y₂−y₁)²]. It comes directly from applying the Pythagoras theorem to the horizontal and vertical gaps between the points.

How do I check if four points form a square using coordinates?

Compute all four side lengths using the distance formula — they must all be equal. Then also check both diagonals are equal in length. Equal sides alone only proves a rhombus; equal diagonals as well confirms a square.

What is the section formula?

The section formula gives the coordinates of a point P that divides the segment joining A(x₁,y₁) and B(x₂,y₂) internally in the ratio m₁:m₂: P = ((m₁x₂+m₂x₁)/(m₁+m₂), (m₁y₂+m₂y₁)/(m₁+m₂)).

What is the midpoint formula?

The midpoint of A(x₁,y₁) and B(x₂,y₂) is ((x₁+x₂)/2, (y₁+y₂)/2) — it's simply the section formula applied with the ratio 1:1.

How do I find the ratio in which a point divides a line segment?

Set up the section formula treating the ratio (m₁:m₂ or k:1) as unknown, using the given point's coordinates. Solve the resulting equation for the ratio, then verify it using the other coordinate as well.

Chapter 7 · CBSE Class 10 Maths

Coordinate Geometry

Two ways to learn — a crisp formula reference, or a story that makes the section formula feel obvious.

Distance formulaShape classificationSection formulaMidpoint formula

Also useful:Chapter 6 — Triangles Chapter 8 — Introduction to Trigonometry NCERT Solutions — Class 10 Maths 1-on-1 CBSE Class 10 Tuition

CBSEClass 10MathematicsChapter 7⏱ 8 min crisp · 11 min story

Four scenes that build the ideas from scratch, step by step.

How this works

Four scenes, one thread: every question about points on a grid — how far apart, what shape, what ratio — turns into ordinary algebra once you have two formulas, both built from ideas already learned (Pythagoras and similar triangles). Read the story, then tap "The Maths" for the formal result.

The Relay Tower

Town B sits 36 km east and 15 km north of Town A. Placing A at the origin, B lands at the point (36, 15). What's the straight-line distance between the towns — without measuring anything on a map?

Engineer

Draw the horizontal leg (36 km) and the vertical leg (15 km) as two sides of a right triangle. The distance between the towns is just the hypotenuse.

By Pythagoras: distance² = 36² + 15² = 1296 + 225 = 1521

distance = √1521 = 39 km

The same trick generalises to ANY two points P(x₁,y₁) and Q(x₂,y₂), not just one on the origin. Draw the horizontal gap (x₂−x₁) and vertical gap (y₂−y₁) as the two legs of a right triangle.

Are They Really a Square?

Champa looks at four classroom desks at A(1,7), B(4,2), C(−1,−1), D(−4,4) and asks Chameli, "Don't you think ABCD is a square?" Chameli isn't so sure.

Chameli

Equal sides alone don't prove it's a square — a rhombus has equal sides too, but its angles aren't 90°. I need to check the diagonals as well.

AB = BC = CD = DA = √34 (all four sides equal — so far, could still be a rhombus)

AC = BD = √68 (both diagonals ALSO equal)

Equal sides AND equal diagonals together are enough — a rhombus with equal diagonals is forced to have 90° angles, making it a genuine square. Chameli was right to insist on checking both.

The same distance formula settles a completely different question too: are three given points even in a straight line? Compute all three pairwise distances — if the two shorter ones add up EXACTLY to the longest, the "triangle" is actually flat, and the points are collinear.

Dividing the Line — The Relay Tower Returns

Back to the towns from Scene 1. A phone company wants a relay tower P between A(0,0) and B(36,15), positioned so its distance from B is exactly twice its distance from A — meaning P splits AB in the ratio 1:2.

Engineer

Draw perpendiculars from P and B down to the x-axis. The two right triangles formed — one small (near P), one large (the whole span to B) — turn out to be similar. That's the AA similarity idea from Chapter 6, showing up again here.

By AA similarity: OD/PC = OP/PB = 1/2 (matches the 1:2 ratio)

This gives: x/(36−x) = 1/2 and y/(15−y) = 1/2

Solving: x = 12, y = 5

The tower belongs at (12, 5). Repeating this same similar-triangles argument for ANY two points A, B and ANY ratio m₁:m₂ gives one general formula — no need to redo the similar-triangle argument by hand every time.

Finding the Ratio, Backwards

Sometimes the question runs in reverse: given the point (−4, 6) and the endpoints A(−6,10), B(3,−8), what ratio does that point actually split AB in?

Student

I don't know the ratio yet — so how do I even start the section formula, which needs the ratio as an input?

Treat the unknown ratio as m₁:m₂ (unknowns), plug the section formula in anyway, and solve algebraically.

−4 = (3m₁ − 6m₂)/(m₁+m₂) → −4m₁−4m₂ = 3m₁−6m₂ → 7m₁ = 2m₂

m₁ : m₂ = 2 : 7

Always double-check using the OTHER coordinate too — plugging m₁:m₂ = 2:7 into the y-coordinate equation should also give exactly 6. If it doesn't, an arithmetic slip happened somewhere.

The same reversed logic solves a related favourite: a parallelogram's diagonals always bisect each other, so the MIDPOINT of one diagonal must equal the midpoint of the other — turning a missing vertex coordinate into one simple equation.

The Big Picture

Every idea in this chapter reduces coordinate-plane geometry to plain algebra: Pythagoras gives the distance formula, similar triangles give the section formula, and both run forwards (find the distance/point) or backwards (find the ratio/missing coordinate) with equal ease.

Distance between two points?

√[(x₂−x₁)²+(y₂−y₁)²]

Is it a square, not just a rhombus?

Equal sides AND equal diagonals

Are three points collinear?

shortest+middle distance = longest

Point dividing AB in ratio m₁:m₂?

Section formula

Ratio given the point?

Solve the section formula for m₁:m₂

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