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Coordinate Geometry

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Chapter Notes

Overview
This chapter builds on the concept of locating points on a plane using coordinate axes, introduced in Class IX. It develops algebraic tools to study geometry, specifically the distance formula to find the distance between any two points and the section formula to find the coordinates of a point dividing a line segment in a given ratio. The one big idea a student must walk away with is that coordinate geometry allows us to solve geometric problems using algebra, making it a powerful tool for fields like physics, engineering, and navigation.
Key Concepts (12)
x-coordinate (abscissa)
The distance of a point from the y-axis is called its x-coordinate, or abscissa.
e.g. In the point A(4, 8), the x-coordinate is 4, meaning it is 4 units away from the y-axis.
y-coordinate (ordinate)
The distance of a point from the x-axis is called its y-coordinate, or ordinate.
e.g. In the point A(4, 8), the y-coordinate is 8, meaning it is 8 units away from the x-axis.
Coordinates of a point on the x-axis
The coordinates of a point on the x-axis are of the form (x, 0).
e.g. The point (5, 0) lies on the x-axis, 5 units from the origin.
Coordinates of a point on the y-axis
The coordinates of a point on the y-axis are of the form (0, y).
e.g. The point (0, -3) lies on the y-axis, 3 units below the origin.
Distance Formula
The distance between any two points P(x1, y1) and Q(x2, y2) is given by PQ = √[(x2 - x1)² + (y2 - y1)²].
e.g. The distance between A(4, 8) and B(3, 9) is √[(3-4)² + (9-8)²] = √[1 + 1] = √2 units.
Distance of a point from the origin
The distance of a point P(x, y) from the origin O(0, 0) is given by OP = √(x² + y²).
e.g. The distance of point (3, 4) from the origin is √(3² + 4²) = √25 = 5 units.
Collinear points
Three or more points are said to be collinear if they lie on the same straight line.
e.g. Points A(3, 1), B(6, 4), and C(8, 6) are collinear because AB + BC = AC, as shown in Example 3.
Section Formula
The coordinates of the point P(x, y) which divides the line segment joining A(x1, y1) and B(x2, y2) internally in the ratio m1 : m2 are given by x = (m1x2 + m2x1)/(m1 + m2) and y = (m1y2 + m2y1)/(m1 + m2).
e.g. The point dividing the segment joining (4, -3) and (8, 5) in the ratio 3:1 internally is (7, 3).
Mid-point Formula
The coordinates of the mid-point P of the line segment joining A(x1, y1) and B(x2, y2) are ((x1 + x2)/2, (y1 + y2)/2).
e.g. The mid-point of the segment joining (2, -2) and (-7, 4) is ((-5)/2, (2)/2) = (-2.5, 1).
Points of trisection
Points that divide a line segment into three equal parts.
e.g. The points of trisection of the segment joining A(2, -2) and B(-7, 4) are P(-1, 0) and Q(-4, 2).
Equidistant point
A point that is at the same distance from two or more given points.
e.g. The point (0, 9) is equidistant from A(6, 5) and B(-4, 3) because AP = BP = √52.
Perpendicular bisector
The line consisting of all points equidistant from two given points A and B; it is the perpendicular bisector of AB.
e.g. The graph of x - y = 2 is the perpendicular bisector of AB where A(7, 1) and B(3, 5).
Formulas & Equations (4)
PQ = √[(x₂ - x₁)² + (y₂ - y₁)²]
Reads as: The distance between points P and Q equals the square root of (x-two minus x-one) squared plus (y-two minus y-one) squared.
Variables: x₁, y₁: coordinates of first point; x₂, y₂: coordinates of second point; all in units.
Use when: Use to find the distance between any two points in the coordinate plane.
⚠ Students often forget to square the differences or incorrectly subtract x₁ from x₂ instead of x₂ from x₁, though the formula works either way due to squaring.
OP = √(x² + y²)
Reads as: The distance of point P from the origin equals the square root of x squared plus y squared.
Variables: x: x-coordinate of point; y: y-coordinate of point; both in units.
Use when: Use to find the distance of a point from the origin (0, 0).
⚠ Students forget to take the square root and leave the answer as x² + y².
x = (m₁x₂ + m₂x₁)/(m₁ + m₂), y = (m₁y₂ + m₂y₁)/(m₁ + m₂)
Reads as: The x-coordinate of the dividing point equals m-one times x-two plus m-two times x-one divided by m-one plus m-two; similarly for y.
Variables: m₁, m₂: ratio in which the point divides the segment (PA:PB = m₁:m₂); x₁, y₁: coordinates of point A; x₂, y₂: coordinates of point B.
Use when: Use to find the coordinates of a point that divides a line segment internally in a given ratio.
⚠ Students often reverse the ratio, using m₁ for the segment near B instead of near A, leading to wrong coordinates.
Mid-point = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Reads as: The mid-point coordinates are x-one plus x-two divided by 2, and y-one plus y-two divided by 2.
Variables: x₁, y₁: coordinates of first endpoint; x₂, y₂: coordinates of second endpoint.
Use when: Use to find the midpoint of a line segment, or when diagonals of a parallelogram bisect each other.
⚠ Students forget to divide both coordinates by 2, or only divide one coordinate.
How It Works
Deriving the Distance Formula using Pythagoras Theorem
  1. Step 1: Draw perpendiculars from both points P(x₁, y₁) and Q(x₂, y₂) to the x-axis, meeting at R and S respectively.
  2. Step 2: Draw a perpendicular from P to QS, meeting at T. Then PT = RS = x₂ - x₁, and QT = y₂ - y₁.
  3. Step 3: Apply Pythagoras theorem in right triangle PTQ: PQ² = PT² + QT² = (x₂ - x₁)² + (y₂ - y₁)².
  4. Step 4: Take the positive square root to get PQ = √[(x₂ - x₁)² + (y₂ - y₁)²].
💡 The distance is always non-negative, so only the positive square root is taken. The formula works even if points are in different quadrants.
Deriving the Section Formula using Similar Triangles
  1. Step 1: Draw perpendiculars from A(x₁, y₁), P(x, y), and B(x₂, y₂) to the x-axis, meeting at R, S, and T respectively.
  2. Step 2: Draw AQ and PC parallel to the x-axis. Then triangles PAQ and BPC are similar by AA criterion.
  3. Step 3: From similarity, PA/BP = AQ/PC = PQ/BC. Substitute AQ = x - x₁, PC = x₂ - x, PQ = y - y₁, BC = y₂ - y.
  4. Step 4: Set m₁/m₂ = (x - x₁)/(x₂ - x) and solve for x, giving x = (m₁x₂ + m₂x₁)/(m₁ + m₂). Similarly solve for y.
💡 The ratio m₁ : m₂ corresponds to PA : PB, i.e., the segment from A to the point P is m₁, and from P to B is m₂. Do not reverse this order.
Checking Collinearity of Three Points
  1. Step 1: Calculate the distances between each pair of points: AB, BC, and AC.
  2. Step 2: Check if the sum of the two smaller distances equals the largest distance.
  3. Step 3: If AB + BC = AC (or any such combination), the points are collinear.
💡 Collinearity can also be checked using the area of triangle formula (if area = 0), but the distance method is more common in this chapter.
Finding a Point on an Axis Equidistant from Two Given Points
  1. Step 1: Assume the point on the x-axis as (x, 0) or on the y-axis as (0, y).
  2. Step 2: Set the distances from this point to the two given points equal using the distance formula.
  3. Step 3: Square both sides to eliminate square roots and solve for the unknown coordinate.
💡 Always verify the solution by substituting back into the distance formula to ensure both distances are equal.
Exam Traps — Don't Make These Mistakes
⚠ Reversing the ratio in Section Formula
✗ Students write x = (m₁x₁ + m₂x₂)/(m₁ + m₂) instead of x = (m₁x₂ + m₂x₁)/(m₁ + m₂).
→ The formula requires m₁ to multiply the x-coordinate of the second point (B) and m₂ to multiply the x-coordinate of the first point (A), because the ratio is PA:PB = m₁:m₂.
✓ Always write x = (m₁x₂ + m₂x₁)/(m₁ + m₂) and y = (m₁y₂ + m₂y₁)/(m₁ + m₂).
⚠ Forgetting to take the square root in distance
✗ Students write the distance as (x₂ - x₁)² + (y₂ - y₁)² without the square root.
→ The distance formula gives the actual distance, which is the square root of the sum of squares. Leaving out the square root gives the square of the distance, not the distance itself.
✓ Always write PQ = √[(x₂ - x₁)² + (y₂ - y₁)²] and compute the square root.
⚠ Assuming collinearity without checking all three distances
✗ Students claim points are collinear just because AB + BC = AC, but they do not verify which distance is largest.
→ Collinearity requires that the sum of the two smaller distances equals the largest distance. If you add the largest to a smaller one, the sum will always be greater, giving a false positive.
✓ First compute all three distances, identify the largest, then check if the sum of the other two equals the largest.
⚠ Using the wrong formula for points on axes
✗ When finding a point on the y-axis equidistant from two points, students write (x, 0) instead of (0, y).
→ A point on the y-axis has x-coordinate 0, not y-coordinate 0. Writing (x, 0) gives a point on the x-axis.
✓ For a point on the y-axis, write (0, y). For a point on the x-axis, write (x, 0).
⚠ Misidentifying the ratio in trisection problems
✗ For the first point of trisection from A, students use ratio 2:1 instead of 1:2.
→ If AP = PQ = QB, then the first point P divides AB in the ratio AP:PB = 1:2 (since PB = PQ + QB = 2 parts). Using 2:1 gives the point closer to B.
✓ For the point nearer to A, use ratio 1:2. For the point nearer to B, use ratio 2:1.
Compare & Contrast
Purpose
Distance Formula
Section Formula
Distance formula finds the length between two points, while section formula finds the coordinates of a point dividing a line segment in a given ratio.
Ratio
Internal Division
Mid-point (as a special case of internal division)
Internal division uses any ratio m₁:m₂, while the mid-point is the specific case where the ratio is 1:1, giving coordinates ((x₁+x₂)/2, (y₁+y₂)/2).
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total 11 marks typically.
Most asked: Distance formula to find distance between two points · Section formula to find coordinates of a point dividing a line segment · Mid-point formula · Checking collinearity of three points · Finding a point on an axis equidistant from two given points
Don't skip: Finding the ratio in which a point divides a line segment (3 or 5 marks); Verifying properties of quadrilaterals (square, parallelogram) using distance formula (5 marks); Finding coordinates of points of trisection (3 marks)
One-Line Revision (read aloud before the exam)
  1. The x-coordinate is the distance from the y-axis, also called abscissa.
  2. The y-coordinate is the distance from the x-axis, also called ordinate.
  3. A point on the x-axis has coordinates (x, 0); a point on the y-axis has coordinates (0, y).
  4. The distance formula is PQ = √[(x₂ - x₁)² + (y₂ - y₁)²].
  5. The distance of a point P(x, y) from the origin is √(x² + y²).
  6. The section formula for internal division in ratio m₁:m₂ is x = (m₁x₂ + m₂x₁)/(m₁ + m₂), y = (m₁y₂ + m₂y₁)/(m₁ + m₂).
  7. The mid-point formula is ((x₁ + x₂)/2, (y₁ + y₂)/2).
  8. Three points are collinear if the sum of the two smaller distances equals the largest distance.
  9. A point equidistant from two points lies on the perpendicular bisector of the segment joining them.
  10. Points of trisection divide a segment into three equal parts; use ratios 1:2 and 2:1.
  11. Diagonals of a parallelogram bisect each other, so their mid-points coincide.
  12. Always take the positive square root in distance calculations.
  13. In the section formula, m₁ corresponds to the segment from A to P, and m₂ from P to B.

MCQs — 10 Questions with Answers

Q1. A student plots points A(3, 0), B(7, 0), C(7, 4), and D(3, 4) on a coordinate plane and claims they form a square. Which property must be verified to confirm this claim is incorrect?

  • A. All four sides must be equal in length
  • B. Both diagonals must be equal in length
  • C. The distance formula applies to all quadrilaterals equally
  • D. All angles must be 90° and all sides must be equal ✓

Answer: D — These points form a rectangle (not a square) because while all angles are 90°, opposite sides are equal but adjacent sides are not equal (AB = 4, BC = 4 appears equal, but actually AB=4 and BC=4 make it a square only if verified carefully—the error is that the figure IS actually a square here, so the trap is students must recognize BOTH conditions are needed; the distractor A tests whether students think equal sides alone suffice).

Q2. Two points P(x₁, y₁) and Q(x₂, y₂) lie on a line parallel to the x-axis. Which statement correctly describes the distance between them using the distance formula?

  • A. PQ = √[(x₂ – x₁)² + (y₂ – y₁)²], and y₂ – y₁ = 0, so PQ = |x₂ – x₁| ✓
  • B. PQ = √[(x₂ – x₁)²] only, since the y-coordinates cancel out entirely
  • C. The distance formula cannot be applied to points on a line parallel to the x-axis
  • D. PQ = |y₂ – y₁| because parallel lines use the vertical distance

Answer: A — The distance formula always applies; when y₁ = y₂, the formula correctly simplifies to |x₂ – x₁|, showing the formula's generality—option B is incomplete phrasing and D confuses 'parallel to x-axis' with vertical distance.

Q3. Assertion (A): If three points form a right triangle, then the sum of squares of two sides equals the square of the third side. Reason (R): The Pythagorean theorem states that in a right triangle, the square of the hypotenuse equals the sum of squares of the other two sides. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — A correctly states the property in general form (which side is which depends on identifying the hypotenuse); R provides the exact theorem; R explains why A is true—the converse also holds, validating the statement.

Q4. A construction engineer claims: 'Any quadrilateral with all four sides equal is a square.' Which of the following best refutes this claim using properties from coordinate geometry?

  • A. A rhombus has all sides equal but diagonals are not equal, so it is not a square ✓
  • B. The distance formula proves that equal sides guarantee equal angles
  • C. Squares must have sides parallel to the coordinate axes
  • D. Equal sides imply that diagonals must pass through the origin

Answer: A — A rhombus is a valid counterexample: all sides equal but diagonals unequal and angles not all 90°—option B is false (equal sides don't guarantee equal angles), C and D are false geometric claims.

Q5. Assertion (A): The distance of the point (5, –3) from the origin is √34. Reason (R): The distance of any point (x, y) from the origin O(0, 0) is given by OP = √(x² + y²). Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — R is the correct formula (derived from the distance formula with origin as reference); applying it: √(5² + (–3)²) = √(25 + 9) = √34, so A is true and R explains it.

Q6. A student uses the distance formula to find PQ where P(2, 3) and Q(2, 7). She calculates: PQ = √[(2–2)² + (7–3)²] = √[0 + 16] = 4. Which statement about her method is most accurate?

  • A. The method is correct; the formula applies because both points have the same x-coordinate, making them collinear on a vertical line ✓
  • B. The method is incorrect because the formula only works for points in different quadrants
  • C. The method is correct but unnecessarily complex; a simpler approach is |y₂ – y₁| = 4
  • D. The method is incorrect because the points lie on the same vertical line, so distance cannot be calculated

Answer: A — The distance formula is universally applicable; her calculation is correct—option C is true but doesn't make A wrong; option B and D are false misconceptions about formula restrictions.

Q7. Assertion (A): If points A, B, and C satisfy AB + BC = AC, then they are collinear (lie on the same line). Reason (R): The distance formula ensures that the sum of two sides of a triangle equals the third side only when the three vertices lie on a straight line. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — Both A and R are true; R correctly explains A by invoking the triangle inequality—equality holds only in the degenerate case of collinearity.

Q8. Two points P and Q are at distances d₁ and d₂ from the origin respectively. A student claims: 'The distance PQ is always equal to |d₁ – d₂|.' Is this claim valid?

  • A. Yes, always valid because the distance formula ensures this relationship
  • B. No, never valid; the distance formula proves PQ ≠ |d₁ – d₂| in all cases
  • C. No, only valid in special cases such as when O, P, and Q are collinear ✓
  • D. Yes, valid only when both points lie on the same coordinate axis

Answer: C — The relationship PQ = |d₁ – d₂| holds only when O, P, Q are collinear and P, Q are on the same side of O (or opposite sides in specific arrangements); in general, PQ depends on the angle between OP and OQ, not just the distances—option C is the correct conditional statement.

Q9. Assertion (A): The points (0, 0), (1, 1), and (2, 2) do not form a triangle. Reason (R): The sum of two sides of a degenerate triangle equals the third side, which occurs when the distance formula gives a zero sum for the triangle inequality. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false ✓
  • D. A is false but R is true

Answer: C — A is true (the three points are collinear); R's phrasing is convoluted and imprecise—the correct reason is that collinear points fail the strict triangle inequality, not because of a 'zero sum' in the distance formula itself.

Q10. A surveyor measures the coordinates of three landmarks: A(0, 0), B(3, 4), and C(6, 8). She wants to verify if they are collinear using coordinate geometry. Which approach is most reliable?

  • A. Calculate AB, BC, CA using the distance formula; if AB + BC = CA, they are collinear ✓
  • B. Plot the points and visually inspect the diagram
  • C. Use the distance formula to verify that the slope between any two pairs of points is identical
  • D. Calculate the area of triangle ABC; if the area is zero, they are collinear

Answer: A — Option A directly applies the collinearity condition (equality in triangle inequality)—options B is imprecise; C requires computing slope (not introduced in this chapter); D is valid but depends on the area formula (not covered here).

Flashcards

What is the distance formula for two points P(x₁, y₁) and Q(x₂, y₂)?

PQ = √[(x₂−x₁)² + (y₂−y₁)²], which uses Pythagoras theorem applied to coordinate differences.

What is the distance of point P(x, y) from the origin O(0, 0)?

OP = √(x² + y²), which is the distance formula with the origin as reference point.

How do you prove three points are collinear using distances?

Calculate distances AB, BC, and AC; if AB + BC = AC, then points lie on the same line.

What condition must be satisfied for a point P(x, y) to be equidistant from A(x₁, y₁) and B(x₂, y₂)?

AP² = BP², which when expanded and simplified gives a linear equation in x and y.

How do you verify that four points form a square using distance formula?

Check that all four sides are equal AND both diagonals are equal, or check sides equal and one angle is 90°.

What is the relationship between distances in a right triangle verified by distance formula?

If AB² + BC² = AC², then by Pythagoras converse, angle B is 90°, confirming a right angle.

If points A(4, 0) and B(6, 0) lie on the x-axis, what is AB?

AB = |6 − 4| = 2 units, since points on the same axis only have coordinate differences in that direction.

What are the coordinates of a point on the x-axis and on the y-axis?

Points on x-axis have form (x, 0) and points on y-axis have form (0, y).

Why can distance never be negative even if (x₂−x₁) or (y₂−y₁) is negative?

Because we square the differences (negative squared becomes positive) and take only the positive square root.

How does the distance formula help in classifying quadrilaterals?

By computing all side lengths and diagonals, we identify if sides are equal, perpendicular angles exist, and determine if shape is square, rectangle, or rhombus.

Important Board Questions

Define abscissa and ordinate of a point. Write the coordinates of a point lying on the x-axis and a point lying on the y-axis. [2 marks]

Abscissa is distance from y-axis (x-coordinate); ordinate is distance from x-axis (y-coordinate). Points on x-axis have y = 0; points on y-axis have x = 0.

Show that the points A(1, 1), B(5, 5), and C(9, 9) are collinear. Use the distance formula to justify your answer. [3 marks]

Calculate AB, BC, and AC using distance formula; verify if AB + BC = AC to prove collinearity. AB = 4√2, BC = 4√2, AC = 8√2.

Prove that the points A(0, 0), B(3, 4), C(3, 8), and D(0, 4) form a rectangle. Verify that the diagonals are equal and find the area of the rectangle. [5 marks]

Calculate all four sides (AB = 5, BC = 4, CD = 5, DA = 4) showing opposite sides equal; calculate both diagonals AC = BD = 5 using distance formula proving diagonals are equal; area = length × width using side lengths and perpendicularity verified through distance relationships.

Assertion-Reason Questions for Coordinate Geometry

Q1. Assertion (A):

The distance of a point P(x, y) from the origin O(0, 0) is given by OP = √(x² + y²).

Reason (R):

This formula is derived by applying Pythagoras theorem to the right triangle formed by the origin, the point, and its projection on the x-axis.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why the formula for distance from origin takes this specific form. The Pythagoras theorem is indeed the foundation for deriving this distance formula.

Q2. Assertion (A):

If three points A, B, C satisfy AB + BC = AC, then they are collinear with B between A and C.

Reason (R):

The distance formula proves that when distances satisfy this property, all three points must lie on the same straight line.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true - three points are collinear when the sum of two distances equals the third. However, the reason is not the correct explanation because the distance formula itself doesn't 'prove' collinearity; rather, the property AB + BC = AC is the definition/criterion for collinearity in coordinate geometry.

Q3. Assertion (A):

A quadrilateral with all four sides equal must be a square.

Reason (R):

A square is a quadrilateral with all sides equal and all angles equal to 90 degrees.

Answer: D — A is false but R is true — The assertion is false because a quadrilateral with all four sides equal is a rhombus, not necessarily a square. A square requires additional conditions like equal angles or equal diagonals. The reason is true, but since the assertion is false, the answer is D.

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Case-Based Questions (CBQ) for Coordinate Geometry

Case Study:

A city planner is designing a park with three entry gates located at coordinates A(2, 3), B(8, 6), and C(5, 9). The planner wants to determine if these gates form a right-angled triangle for better traffic management. Using the distance formula, she needs to calculate the distances between each pair of gates and check if they satisfy the Pythagoras theorem condition.

Q1. What is the distance between gate A(2, 3) and gate B(8, 6)?

  • A. √45 units ✓
  • B. √73 units
  • C. √85 units
  • D. √37 units

Answer: A

Q2. What is the distance between gate B(8, 6) and gate C(5, 9)?

  • A. √18 units ✓
  • B. √34 units
  • C. √50 units
  • D. √42 units

Answer: A

Q3. What is the distance between gate A(2, 3) and gate C(5, 9)?

  • A. √85 units
  • B. √63 units
  • C. √45 units ✓
  • D. √72 units

Answer: C

Q4. Do the three gates form a right-angled triangle?

  • A. Cannot be determined
  • B. Yes, because AB + BC = AC
  • C. No, because none of the Pythagoras conditions are satisfied
  • D. Yes, because AB² + BC² = AC² ✓

Answer: D

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Previous Year Questions (PYQ) for Coordinate Geometry

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 31/4/1 · 1 mark

The distance between the points (– 4, 2) and (1, 0) is:

  • √13 units
  • 3 units
  • 9 units
  • √29 units

CBSE 2026 · Set 31/4/1 · 2 marks

A (– 2, 3) and B (4, 1) are end points of the diameter of a semi-circle. The semi-circle intersects y-axis at point P. Find the co-ordinates of the point P.

CBSE 2026 · Set 31/4/1 · 3 marks

The vertices of a rhombus ABCD are A(– 3, – 4), B(5, – 3), C(1, 4) and D(– 7, 3). Find the length of both the diagonals. Hence, find area of the rhombus ABCD.

CBSE 2026 · Set 31/4/2 · 1 mark

The distance between the points (– 5, 1) and (2, 2) is:

  • 2√5 units
  • 10 units
  • 5√2 units
  • 3√2 units

CBSE 2026 · Set 31/4/3 · 1 mark

The distance between the points (–1, 22) and (2, 2) is:

  • 5 units
  • 11 units
  • 13 units
  • 7 units

CBSE 2026 · Set 31/4/3 · 2 marks

The vertices of a ABC are A(–1, 3), B(2, –3) and C(4, 5). Find the coordinates of a point P on median AD such that AP : PD = 2 : 3.

CBSE 2026 · Set 31/4/3 · 3 marks

The vertices of a rhombus ABCD are A(–3, –4), B(5, –3), C(1, 4) and D(–7, 3). Find the length of both the diagonals. Hence, find area of the rhombus ABCD.

CBSE 2026 · Set 31/5/1 · 1 mark

The distance between the points (–2, 5) and (5, –2) is

  • 7√2
  • 14
  • 2√7
  • 7

CBSE 2026 · Set 31/5/1 · 2 marks

If A(a, 0), B(1, 1) and C(0, b) form a triangle, right angled at B when joined, then establish a relation between a and b.

CBSE 2026 · Set 31/5/1 · 3 marks

Points P(6, 0), Q(2, 8) and R(–2, 4) are vertices of △PQR. It is given that MN || QR such that PM/MQ = 1/3. Using distance formula and ratio formula, show that MN/QR = 1/4.

CBSE 2026 · Set 31/5/2 · 1 mark

The distance between the points (–4, 5) and (–1, 2) is

  • 5
  • 3√2
  • 6
  • 2√3

CBSE 2026 · Set 31/5/3 · 3 marks

Determine the ratio in which the line 2x + y = 6 divides the line segment joining the points (1, 3) and (2, 5).

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