The distance between the points \((-3, 4)\) and \((5, -2)\) is
- (a)\(8\)
- (b)\(14\)
- (c)\(\sqrt{28}\)
- (d)\(10\)
Coordinate Geometry Class 10: notes and important questions
Revision notes, 40 board-style questions with the step marks shown, and a four-step route from the basics to 95+, each step ending in a short checkpoint.
- 40 questions
- 12 multiple choice, 3 assertion–reason, 8 very short answer, 9 short answer, 5 long answer, 3 case study
- About 9 hours to master
Coordinate Geometry unit: 6 of 80 theory marks (Coordinate Geometry is the only chapter in this unit).
Revision notes
Coordinate Geometry — revision notes
1. Distance formula
\[PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
- Distance from the origin: \(\sqrt{x^2 + y^2}\). Distance from the \(x\)-axis is \(|y|\); from the \(y\)-axis is \(|x|\).
- Points on the \(x\)-axis are \((x, 0)\); points on the \(y\)-axis are \((0, y)\).
2. Section formula (internal division)
The point dividing the segment joining \(A(x_1, y_1)\) and \(B(x_2, y_2)\) internally in the ratio \(m_1 : m_2\) is \[\left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2},\ \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right)\]
- Finding an unknown ratio: take it as \(k : 1\) and use whichever coordinate is known (e.g. \(y = 0\) for the \(x\)-axis).
- Trisection points: ratios \(1 : 2\) and \(2 : 1\). Four equal parts: mid-point, then mid-points of each half.
3. Mid-point formula
\[M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\]
4. Identifying shapes
- Collinear: sum of two distances equals the third.
- Right triangle: \(a^2 + b^2 = c^2\) for the side lengths (converse of Pythagoras, known from earlier classes).
- Parallelogram: diagonals bisect each other (same mid-point). Rhombus: all sides equal. Rectangle: opposite sides equal and diagonals equal. Square: all sides equal and diagonals equal.
Worked example 1
Distance between \((2, -1)\) and \((-4, 7)\): \(\sqrt{36 + 64} = 10\).
Worked example 2
Point dividing \(A(1, 7)\), \(B(9, -5)\) in \(3 : 1\): \(\left(\dfrac{27 + 1}{4}, \dfrac{-15 + 7}{4}\right) = (7, -2)\).
Worked example 3
Ratio in which the \(y\)-axis divides \(P(-2, 3)\), \(Q(6, -1)\): \(\dfrac{6k - 2}{k + 1} = 0 \Rightarrow k = \dfrac13\), ratio \(1 : 3\); point \((0, 2)\).
Common errors
- Swapping \(m_1\) and \(m_2\) in the section formula — \(m_1\) multiplies the coordinates of the second point.
- Forgetting the square root, or squaring a negative difference incorrectly.
- Concluding “square” from equal sides alone (a rhombus also has equal sides); check the diagonals.
- Leaving the final distance unsimplified (\(\sqrt{72} = 6\sqrt2\)).
Board-exam tips
- Write the formula first, then substitute — method marks are given for correct substitution.
- For shape proofs, compute all four sides and both diagonals and then state the conclusion with a reason.
- Note: area of a triangle using coordinates is not in the current syllabus.
Topics in this chapter: Distance formula · Mid-point formula · Section formula · Applications to geometric shapes.
Route to 95: four steps
Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.
Step 1
Secure the basics
You can find the distance between two points and the mid-point of a segment without slips.
Step 2
Board standard
You can use the section formula both ways: to find the dividing point and to find the ratio (or an unknown coordinate).
Step 3
Full marks on long answers
You can prove the type of a quadrilateral or triangle from coordinates in a full, reasoned long answer.
Step 4
95+ stretch (HOTS)
You can handle trisection points, points equidistant from three given points and right-angle conditions.
Practice questions
Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 4 of the 40 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.
The distance of the point \((5, -12)\) from the origin is
- (a)\(13\)
- (b)\(17\)
- (c)\(7\)
- (d)\(\sqrt{119}\)
If the distance between \((4, p)\) and \((1, 0)\) is \(5\), then \(p\) equals
- (a)\(4\) only
- (b)\(\pm 3\)
- (c)\(-4\) only
- (d)\(\pm 4\)
Where marks are lost in Coordinate Geometry
- Swapping m₁ and m₂ in the section formula. Fix: m₁ multiplies the coordinates of the second point; write ((m₁x₂ + m₂x₁)/(m₁ + m₂), …) before substituting.
- Calling a quadrilateral a square from four equal sides. Fix: compute all four sides and both diagonals; equal sides and equal diagonals give a square, equal sides alone a rhombus.
- Squaring negative differences wrongly or forgetting the square root. Fix: write (x₂ − x₁)² with brackets and take the root on the last line.
- Leaving surds unsimplified (√72 instead of 6√2). Fix: simplify at the end; some schemes award the A1 only for the simplified form.
- For 'ratio in which the y-axis divides', not using x = 0 (or y = 0 for the x-axis). Fix: take the ratio as k : 1 and set the relevant coordinate to zero.
- Wasting time on area of a triangle by coordinates. Fix: it is deleted from the current syllabus; use distances and Pythagoras instead.
Coordinate Geometry in our sample papers
Once step 3 is passed, test the chapter inside a full timed paper on the 2026-27 pattern (original papers by us, not official CBSE papers).
- Sample paper 1 (Board standard): questions 6, 7, 37
- Sample paper 2 (Board standard): questions 7, 8, 20, 28
- Sample paper 3 (Above board): questions 8, 9, 37
- Sample paper 4 (Above board): questions 8, 23, 28
- Sample paper 5 (95+ challenge): questions 7, 8, 37