Class 10 Coordinate Geometry Notes: Distance Formula, Section Formula and Board Questions

Table of Contents

1. Why Coordinate Geometry Is Quick Marks
2. The Basics: Points, Axes and Quadrants
3. The Distance Formula
4. The Section Formula
5. Midpoints, Trisection Points and Shapes
6. Common Mistakes and a Final Checklist
7. FAQs

Coordinate Geometry in Class 10 comes down to two formulas. The distance formula tells you how far apart two points are. The section formula tells you where a point lies on the line joining two others. Almost every question in the chapter uses one or both.

That makes it one of the quickest chapters to prepare. There is very little to memorise, the methods are mechanical, and the answers are easy to check. The marks go to students who apply the formulas carefully and avoid a few predictable slips.

These notes cover the whole chapter in the order it is tested, with worked examples, the question types the board likes, and the mistakes that cost marks. For complete chapter notes with practice questions, see ZetaGalaxy Class 10 notes.

Why Coordinate Geometry Is Quick Marks

ZetaGalaxy mapped past CBSE Class 10 maths board papers question by question. In that analysis, Coordinate Geometry ranked seventh among all fourteen chapters, and in the 2023 papers it was among the five most-asked chapters of the year, with 7.5 percent of all questions.

Inside the chapter, the Section Formula was the standout topic. It was asked in 253 questions across the papers analysed, carrying 571 marks. That puts it among the ten most-asked topics in the whole Class 10 maths paper, a strong result for a single formula.

The syllabus gives Coordinate Geometry 6 marks as a unit, but its formulas also appear inside other questions. Finding a midpoint, checking whether three points are collinear, or proving that a shape is a particular quadrilateral all draw on the same two tools.

The question types are steady. Short questions ask for a distance, a midpoint or a missing coordinate. Medium questions ask for the point dividing a line in a given ratio, or the ratio in which an axis divides a line. Longer questions and case studies place points on a map, a classroom layout or a garden plan and ask you to work with them.

The Basics: Points, Axes and Quadrants

Every point in the plane is written as an ordered pair (x, y). The first number, x, is the abscissa, the distance from the y-axis. The second, y, is the ordinate, the distance from the x-axis.

A few facts are asked as 1-mark questions:

• Any point on the x-axis has y = 0, so it looks like (x, 0).
• Any point on the y-axis has x = 0, so it looks like (0, y).
• The origin is (0, 0).
• In the first quadrant, both coordinates are positive. In the second, x is negative and y is positive. In the third, both are negative. In the fourth, x is positive and y is negative.

These facts look simple, and they are where several short questions in the chapter begin. A question that says a point lies on the x-axis is quietly telling you that its y-coordinate is zero. Spotting that clue often turns a two-unknown problem into a one-unknown problem.

The Distance Formula

The distance between two points A(x₁, y₁) and B(x₂, y₂) is:

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

It is just the Pythagoras theorem. The difference in x is one side of a right triangle, the difference in y is the other, and the distance is the hypotenuse. If you understand that, you can rebuild the formula any time.

The distance of a point (x, y) from the origin is √(x² + y²).

Take a simple example. The distance between (2, 3) and (4, 1) is √[(4 − 2)² + (1 − 3)²] = √(4 + 4) = √8 = 2√2.

The distance formula is used in three common question types.

Finding an unknown coordinate. Find the point on the x-axis that is equidistant from (2, −5) and (−2, 9). A point on the x-axis is (x, 0). Set the two distances equal and square both sides to remove the roots: (x − 2)² + 25 = (x + 2)² + 81. Expanding, x² − 4x + 4 + 25 = x² + 4x + 4 + 81. The x² terms cancel, leaving −8x = 56, so x = −7. The point is (−7, 0). To check, its distance squared from each point is 81 + 25 = 106 and 25 + 81 = 106.

Checking collinearity. Three points A, B and C lie on one line if the two shorter distances add up to the longest one, such as AB + BC = AC.

Try A(1, 5), B(2, 3) and C(−2, 11). AB = √(1 + 4) = √5. AC = √(9 + 36) = √45 = 3√5. BC = √(16 + 64) = √80 = 4√5. The two shorter distances add up to √5 + 3√5 = 4√5, which is exactly BC. So the three points are collinear. Simplifying each root into a multiple of √5 is what makes the comparison easy, so always simplify before you add.

Identifying a shape. Find all the side lengths, and the diagonals if needed. Four equal sides and equal diagonals make a square. Four equal sides with unequal diagonals make a rhombus. Opposite sides equal with equal diagonals make a rectangle. Write each distance clearly and state the conclusion in words.

The Section Formula

The section formula is the most-asked topic in the chapter. The point P that divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m : n is:

P = ( (mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n) )

The pattern to remember is that m multiplies the coordinates of B, the far point, and n multiplies the coordinates of A. Writing it the other way round is the single most common error in this chapter.

Find the point dividing the segment joining (−1, 7) and (4, −3) in the ratio 2 : 3. Here m = 2, n = 3. x = (2 × 4 + 3 × (−1))/5 = (8 − 3)/5 = 1. y = (2 × (−3) + 3 × 7)/5 = (−6 + 21)/5 = 3. The point is (1, 3).

The board also asks the reverse question. Find the ratio in which the y-axis divides the segment joining (5, −6) and (−1, −4). Let the ratio be k : 1. Any point on the y-axis has x = 0, so (k × (−1) + 1 × 5)/(k + 1) = 0. That gives −k + 5 = 0, so k = 5. The ratio is 5 : 1. Putting k = 5 into the y-coordinate gives (5 × (−4) + 1 × (−6))/6 = −26/6 = −13/3. The point of division is (0, −13/3).

Using k : 1 instead of m : n is the trick for this type. It leaves only one unknown, which one condition is enough to find. If k comes out negative, the point divides the segment externally, which the Class 10 questions do not ask for, so recheck your working.

Midpoints, Trisection Points and Shapes

The midpoint is the special case of the section formula where the ratio is 1 : 1. The midpoint of A(x₁, y₁) and B(x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2).

The midpoint formula is behind a favourite question about parallelograms. The diagonals of a parallelogram bisect each other, so they share the same midpoint. If three vertices are given, set the midpoint of one diagonal equal to the midpoint of the other, and solve for the missing vertex.

Trisection points divide a segment into three equal parts. They are found with the section formula in the ratios 1 : 2 and 2 : 1. For the segment joining (4, −1) and (−2, −3), the ratio 1 : 2 gives ((−2 + 8)/3, (−3 − 2)/3) = (2, −5/3). The ratio 2 : 1 gives ((−4 + 4)/3, (−6 − 1)/3) = (0, −7/3). These are the two points of trisection.

Case studies often use these ideas in a real setting. Plants placed along a line in a garden, students standing in a classroom grid, or towns marked on a map. Read the setting, write down each coordinate, and the question becomes one of the types above.

Common Mistakes and a Final Checklist

The mistakes in this chapter are mostly about care, not understanding:

• Swapping m and n in the section formula. m goes with the second point.
• Sign errors when a coordinate is negative. Put brackets around every negative value before squaring or multiplying.
• Forgetting to square both sides when setting two distances equal, or leaving a square root in the answer that could be simplified.
• Missing the clue that a point on an axis has one coordinate equal to zero.
• Naming a shape from its sides alone without checking the diagonals.
• Giving coordinates without stating the final point clearly as an ordered pair.

A quick checklist before the exam:

• Can you write the distance formula and explain it as Pythagoras?
• Can you find an unknown point that is equidistant from two others?
• Can you apply the section formula with m and n in the right places?
• Can you find the ratio in which an axis divides a segment using k : 1?
• Can you use midpoints to find the fourth vertex of a parallelogram, and find trisection points?

If yes to all five, you have covered almost every mark this chapter carries.

ZetaGalaxy's chapter notes for Coordinate Geometry include practice by marks and board-pattern questions. If a formula or a case study does not make sense, Nova, the AI teacher, walks through it at your level.

Conclusion

Coordinate Geometry is two formulas used carefully. The section formula alone carried 571 marks across the papers in ZetaGalaxy's analysis, and the distance formula sits behind most of the rest. Learn both, practise the five question types above, and guard against sign errors and swapped ratios.

Download the complete chapter notes at ZetaGalaxy Class 10 notes and work through the practice questions.

FREQUENTLY ASKED QUESTIONS

Q: What is the most important topic in Class 10 Coordinate Geometry?

A: The section formula. In ZetaGalaxy's analysis of past board papers, it was asked in 253 questions carrying 571 marks, making it one of the ten most-asked topics in Class 10 maths.

Q: What is the section formula?

A: The point dividing the join of (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)). m multiplies the second point's coordinates.

Q: How do I find the ratio in which an axis divides a line segment?

A: Take the ratio as k : 1 and apply the section formula. On the x-axis, set the y-coordinate to zero. On the y-axis, set the x-coordinate to zero. Solve for k.

Q: How do I check if three points are collinear?

A: Find the three distances between them. If the two shorter distances add up to the longest, the points lie on one line.

Q: Where can I download Class 10 Coordinate Geometry notes?

A: You can download the chapter notes from zetagalaxy.com/notes/class-10, along with notes for the other Class 10 maths chapters. -