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Coordinate geometry turns geometry into arithmetic. Once points have coordinates, you can find the length of a side or the point that splits a segment in a given ratio without drawing anything to scale. Class 10 keeps this to two formulae, and most lost marks come from signs and the order of a ratio. This post covers what is in the syllabus, works four examples, and lists the slips to avoid.
What is in the Class 10 syllabus?
CBSE's Class X Mathematics syllabus for 2026-27 lists a review of coordinate geometry, the distance formula and the "section formula (internal division)". The unit is worth 6 of the 80 theory marks. The NCERT chapter, rationalised and reprinted for 2026-27, has only these two sections. Older books also had the area of a triangle from coordinates; it is not in the current chapter, so check your syllabus before practising it.
What are the two formulae?
Distance formula. The distance between P(x₁, y₁) and Q(x₂, y₂) is:
PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]
It is the Pythagoras theorem on the right triangle whose legs are the horizontal and vertical gaps between the points.
Section formula. The point P that divides the segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m : n is:
P = ( (m x₂ + n x₁)/(m + n), (m y₂ + n y₁)/(m + n) )
The midpoint is the special case m = n = 1: ((x₁ + x₂)/2, (y₁ + y₂)/2).
Worked examples
Example 1. Distance. Find the distance between A(−2, 3) and B(4, −5).
x₂ − x₁ = 4 − (−2) = 6. y₂ − y₁ = −5 − 3 = −8.
AB = √(6² + (−8)²) = √(36 + 64) = √100 = 10 units.
Example 2. Section formula. Find the point P that divides AB in the ratio 1 : 3, for the same A and B.
m = 1, n = 3.
x = (1 × 4 + 3 × (−2))/4 = (4 − 6)/4 = −1/2.
y = (1 × (−5) + 3 × 3)/4 = (−5 + 9)/4 = 1.
P = (−1/2, 1).
Check: P should be a quarter of the way from A to B. A quarter of the x-gap is 6/4 = 1.5, and −2 + 1.5 = −0.5. A quarter of the y-gap is −8/4 = −2, and 3 − 2 = 1. It matches.
Example 3. Finding the ratio. In what ratio does the x-axis divide AB, with A(−2, 3) and B(4, −5)? Find the point too.
Let the ratio be k : 1. Any point on the x-axis has y = 0.
(k × (−5) + 1 × 3)/(k + 1) = 0, so −5k + 3 = 0 and k = 3/5.
The ratio is 3 : 5.
x = (3 × 4 + 5 × (−2))/8 = (12 − 10)/8 = 1/4. The point is (1/4, 0).
Example 4. Collinear points. Are A(1, 1), B(3, 5) and C(4, 7) collinear?
AB = √(2² + 4²) = √20 = 2√5.
BC = √(1² + 2²) = √5.
AC = √(3² + 6²) = √45 = 3√5.
AB + BC = 2√5 + √5 = 3√5 = AC. So B lies on AC and the points are collinear.
Why mix chapters when you practise?
A chapter exercise tells you the method before you start. A board paper does not. In a cluster randomised trial across 54 seventh-grade maths classes, Rohrer and colleagues (2020) found that classes doing interleaved assignments, where no two consecutive problems needed the same strategy, scored 61% on an unannounced test a month later, against 38% for classes doing blocked practice. Shuffle distance, section and other chapters' problems so that you practise choosing the formula.
Which mistakes lose marks?
Losing a minus sign. Write 4 − (−2), not 4 − 2. Put every negative coordinate in brackets when you substitute.
m and n swapped. See the callout. Check that the point lies where the ratio says it should.
Forgetting the square root. The distance is √100 = 10, not 100.
Not simplifying surds. √20 should become 2√5 so that sums like Example 4 are easy to see.
Leaving k as the answer. Example 3 asks for a ratio and a point. State "3 : 5" and the coordinates.
Using the midpoint formula for other ratios. It only works for 1 : 1.
When a problem stalls because you cannot see which formula fits, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can see how the set-up was chosen.
A one-week practice routine
Day 1: Six distance problems, half with negative coordinates. Simplify every surd.
Day 2: Five section formula problems. Check each answer with the "fraction of the gap" method from Example 2.
Day 3: Four "find the ratio" problems using k : 1, including ones where the x-axis or y-axis does the dividing.
Day 4: Three proof-style problems: collinear points, an isosceles triangle, or a parallelogram whose diagonals bisect each other.
Day 5: A mixed set of eight, shuffled with problems from triangles and trigonometry.
When you reach Class 11, the same ideas continue in straight lines in coordinate geometry. For neat working that catches sign slips, read how neat working reduces errors.
Today, solve Example 2 again with the ratio 3 : 1 and check that your point is now closer to B.
