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§7Class 10, Chapter 7

Coordinate Geometry: Distance, Ratios, and Where Points Meet

Coordinate geometry turns location and distance into pure algebra. This companion covers the distance formula and the section formula, from GPS-style distance checks to finding the exact point that splits a segment in any ratio, with tools, flashcards and a CBSE-style quiz to test yourself.

d=√[(x₂−x₁)²+(y₂−y₁)²]section formula(x, y)midpoint

📍7.1 Locating Points: A Quick Recap

Back in Class IX you learned that any point on a flat surface can be pinned down with just two numbers: its x-coordinate (also called the abscissa), which is its distance from the y-axis, and its y-coordinate (also called the ordinate), which is its distance from the x-axis. Points sitting exactly on the x-axis always look like (x, 0), and points sitting exactly on the y-axis always look like (0, y).

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Think of it like a map pin.. A GPS pin only ever needs two numbers, latitude and longitude, to tell you exactly where something is on a flat map. A coordinate pair (x, y) does the exact same job on the coordinate plane — no more, no less information is needed.

You've already met two big ideas that connect algebra and geometry: a linear equation ax + by + c = 0 always graphs as a straight line, and a quadratic equation y = ax² + bx + c graphs as a parabola. Coordinate geometry is the toolkit that makes this connection work in both directions — it lets you study shapes using algebra, and understand algebra using shapes.

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Where this shows up in the real world.. Coordinate geometry isn't just a classroom exercise — it's the working language behind physics simulations, engineering blueprints, ship and aircraft navigation, earthquake (seismology) tracking, and even digital art and animation, anywhere a computer needs to know exactly where something is and how far it is from something else.

🎯What This Chapter Covers

  • The Distance Formula — how to find the straight-line distance between any two points, given only their coordinates.
  • The Section Formula — how to find the exact coordinates of a point that divides a line segment joining two given points in a given ratio.

📏7.2 The Distance Formula: How Far Apart Are Two Points?

Imagine a town B that sits 36 km east and 15 km north of town A. You want to know the straight-line distance from A to B — the distance a bird would fly, or a drone would travel, without following any roads. If you place A at the origin, B ends up at the point (36, 15), and the segment AB becomes the hypotenuse of a right triangle. That means you can reach for the Pythagoras Theorem.

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"As the crow flies.". This is exactly the phrase people use for straight-line distance — ignoring roads, rivers, or detours. A delivery drone flying directly from a warehouse to a customer's rooftop is measuring distance exactly the way this section teaches you to.

➖The Easy Cases: Points on the Same Axis

If two points both sit on the x-axis, say A(4, 0) and B(6, 0), the distance between them is simply the difference of their x-coordinates: AB = OB − OA = 6 − 4 = 2 units. The same logic works for two points on the y-axis: for C(0, 3) and D(0, 8), CD = 8 − 3 = 5 units.

Points on x-axis: AB = |x₂ − x₁|
Points on y-axis: CD = |y₂ − y₁|

But what about a point on the x-axis and a point on the y-axis, like A(4, 0) and C(0, 3)? Now OA = 4 and OC = 3 are two legs of a right angle at the origin, so AC = √(3² + 4²) = 5 units — Pythagoras again. This hints at the general method: whenever two points don't share an axis, draw perpendiculars to build a right triangle, then let Pythagoras do the work.

🧮Building the General Formula

Now take any two points P(x₁, y₁) and Q(x₂, y₂), anywhere on the plane. Drop perpendiculars from P and Q down to the x-axis, meeting it at R and S. Draw one more perpendicular from P across to meet QS at a point T. Triangle PTQ is now a right triangle with its right angle at T, and PQ is its hypotenuse.

SET UPOR = x₁ and OS = x₂, so the horizontal leg RS = x₂ − x₁. Since PRST is a rectangle, PT = RS = x₂ − x₁.
VERTICAL LEGSQ = y₂ and ST = PR = y₁, so the vertical leg QT = y₂ − y₁.
APPLY PYTHAGORASIn right triangle PTQ: PQ² = PT² + QT² = (x₂ − x₁)² + (y₂ − y₁)².
CONCLUDETaking the positive square root (distance is never negative): PQ = √[(x₂ − x₁)² + (y₂ − y₁)²].
📐The Distance Formula. The distance between P(x₁, y₁) and Q(x₂, y₂) is PQ = √[(x₂ − x₁)² + (y₂ − y₁)²].
PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]
⚠️The subtraction order doesn't matter.. Since every difference gets squared, (x₁ − x₂)² is always identical to (x₂ − x₁)². So PQ = √[(x₁ − x₂)² + (y₁ − y₂)²] is exactly the same formula written the other way around — pick whichever order feels natural.

🎯Special Case: Distance From the Origin

If one of the two points is the origin O(0, 0), the formula collapses to something simpler and very useful.

OP = √(x² + y²) for a point P(x, y)

✏️Worked Examples

Check whether P(3, 2), Q(−2, −3) and R(2, 3) form a triangle, and identify the type.

COMPUTE PQPQ = √[(3+2)² + (2+3)²] = √(25+25) = √50 ≈ 7.07
COMPUTE QRQR = √[(−2−2)² + (−3−3)²] = √(16+36) = √52 ≈ 7.21
COMPUTE PRPR = √[(3−2)² + (2−3)²] = √(1+1) = √2 ≈ 1.41
TRIANGLE CHECKThe sum of any two of these distances exceeds the third, so P, Q, R genuinely form a triangle.
RIGHT-ANGLE CHECKPQ² + PR² = 50 + 2 = 52 = QR². By the converse of Pythagoras, ∠P = 90°, so PQR is a right triangle.

➗7.3 The Section Formula: Finding a Point In Between

Suppose a telephone company wants to place a relay tower at point P somewhere between two towns A and B, such that the tower's distance from B is exactly twice its distance from A. That means P divides segment AB in the ratio 1 : 2. If A is the origin and B is at (36, 15), where exactly should the tower go?

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Like scaling a recipe.. Think of a point dividing a segment in a ratio like sharing a cake unevenly between two people, one gets a bigger share, one gets a smaller share, but the total is fixed. The section formula tells you exactly where that "weighted" splitting point lands.

🧮Building the General Formula

Let A(x₁, y₁) and B(x₂, y₂) be any two points, and let P(x, y) divide AB internally so that PA : PB = m₁ : m₂. Drawing perpendiculars from A, P and B to the x-axis (and a helper line AQ parallel to it) creates two similar triangles, △PAQ and △BPC, by the AA similarity criterion.

SET UP SIMILAR TRIANGLES△PAQ ~ △BPC gives PA/BP = AQ/PC = PQ/BC.
IDENTIFY THE LEGSAQ = x − x₁, PC = x₂ − x, PQ = y − y₁, BC = y₂ − y.
SOLVE FOR xFrom m₁/m₂ = (x − x₁)/(x₂ − x), cross-multiplying and rearranging gives x = (m₁x₂ + m₂x₁) / (m₁ + m₂).
SOLVE FOR yThe same steps with the y-coordinates give y = (m₁y₂ + m₂y₁) / (m₁ + m₂).
📐The Section Formula. The point P(x, y) which divides the segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ has coordinates ( (m₁x₂ + m₂x₁)/(m₁+m₂), (m₁y₂ + m₂y₁)/(m₁+m₂) ).
x = (m₁x₂ + m₂x₁) / (m₁ + m₂)
y = (m₁y₂ + m₂y₁) / (m₁ + m₂)

🔢Writing the Ratio as k : 1

The ratio m₁ : m₂ can always be rewritten as k : 1, where k = m₁/m₂. This is often more convenient for algebra, since there's only one unknown to solve for.

x = (kx₂ + x₁) / (k + 1)
y = (ky₂ + y₁) / (k + 1)

🎯Special Case: The Midpoint Formula

The midpoint of a segment is simply the point that divides it in the ratio 1 : 1. Substituting m₁ = m₂ = 1 into the section formula makes both denominators 2 and gives a clean average of the coordinates.

📍The Midpoint Formula. The midpoint of the segment joining P(x₁, y₁) and Q(x₂, y₂) is ( (x₁+x₂)/2, (y₁+y₂)/2 ).
Midpoint = ( (x₁+x₂)/2, (y₁+y₂)/2 )
⚠️Order matters here — unlike the distance formula.. In PA : PB = m₁ : m₂, m₁ always corresponds to the share measured from A, and m₂ to the share measured from B. Swap the ratio around by mistake and you'll compute the point that's the mirror image along the segment, a genuinely different point.

✏️Worked Examples

Find the coordinates of the point dividing the segment joining (4, −3) and (8, 5) in the ratio 3 : 1, internally.

APPLY FORMULAx = [3(8) + 1(4)] / (3+1) = 28/4 = 7
APPLY FORMULAy = [3(5) + 1(−3)] / (3+1) = 12/4 = 3
CONCLUDEThe required point is (7, 3).
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Where the section formula shows up.. Computer graphics use the section formula constantly to interpolate points along a line (drawing smooth curves and animations frame by frame), and physics uses the same midpoint logic to locate the centre of mass of a uniform rod.

📝7.4 Chapter Summary

  1. Distance Formula: the distance between P(x₁, y₁) and Q(x₂, y₂) is PQ = √[(x₂−x₁)² + (y₂−y₁)²].
  2. Distance from the Origin: the distance of P(x, y) from O(0, 0) is OP = √(x² + y²).
  3. Section Formula: the point P(x, y) dividing the segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ is ( (m₁x₂+m₂x₁)/(m₁+m₂), (m₁y₂+m₂y₁)/(m₁+m₂) ).
  4. Midpoint Formula: the midpoint of the segment joining P(x₁, y₁) and Q(x₂, y₂) is ( (x₁+x₂)/2, (y₁+y₂)/2 ).
📖A note for later.. Everything in this chapter covers internal division, where P lies between A and B on the segment itself. If P lies on the line AB but outside the segment, with PA : PB = m₁ : m₂, that's called external division — a case you'll meet with its own formula in higher classes.