Distance Formula Class 9 ICSE — Maths Chapter 28
The distance formula class 9 ICSE chapter shows how coordinate geometry measures the gap between two points on the Cartesian plane. This Mathematics chapter 28 builds the formula from Pythagoras’ theorem, then applies it to the origin, the axes, collinearity, shape proofs, circle area, and the circumcentre of a triangle. Work through each section with the chapter open, and the exercise questions become straightforward.
Download Distance Formula Class 9 ICSE Mathematics Chapter 28 PDF
Download the Distance Formula Class 9 ICSE Mathematics chapter PDF to revise offline or print for practice. It is the full chapter as printed in the book — 6 pages, 3 sections.
What is inside Chapter 28
6 pages · 3 sections · 10 worked examples
| Section | Title |
|---|---|
| 28.1 | INTRODUCTION |
| 28.2 | THE DISTANCE FORMULA |
| 28.3 | CIRCUMCENTRE OF A TRIANGLE |
28.1 INTRODUCTION
Coordinate geometry works with two known points on the Cartesian plane. Given those points, the chapter lists five problems you can solve. It can find the distance between the points, the coordinates of a point dividing the line in a given ratio, the midpoint of the segment, the equation of the straight line through the points, and the equation of the perpendicular bisector of that line.
This list sets the scope of the chapter. The distance formula chapter in Class 9 ICSE Mathematics focuses on the first item. Every later tool — midpoint, ratio division, line equations — builds on measuring distance first. Keep the five-part list in mind as you read.
The formula you will meet in section 28.2 is the foundation for all of these. The official CISCE syllabus lists coordinate geometry as a core Class 9 topic, so mastering this one formula carries through the rest of the unit. You can also browse the full Class 9 Maths book hub to see how this chapter connects to the course.
28.2 THE DISTANCE FORMULA
Deriving the formula from Pythagoras’ theorem
Let A (x_1, y_1) and B (x_2, y_2) be two points. Draw AC parallel to the x-axis and BC parallel to the y-axis. Then AC equals the difference of the abscissae, x_2 - x_1, and BC equals the difference of the ordinates, y_2 - y_1. Triangle ABC is right-angled at C, so Pythagoras applies.
AB^2 = AC^2 + BC^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2
Taking the positive square root gives the Class 9 ICSE distance formula:
AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
Since (a - b)^2 = (b - a)^2, the order of subtraction does not matter. The formula holds in every quadrant, as long as the proper signs of the coordinates are used. For the points (−4, 7) and (2, −1):
\sqrt{(2 - (-4))^2 + (-1 - 7)^2} = \sqrt{36 + 64} = \sqrt{100} = 10
The distance is 10 units.
Distance from the origin
For any point (x, y), the origin is (0, 0). Substituting into the formula gives a shorter version:
\sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}
For (9, 12), the distance is \sqrt{81 + 144} = 15 units. This form appears in many exercise questions, so learn it alongside the full formula.
Points on the axes
A point on the x-axis has ordinate zero and is written (x, 0). A point on the y-axis has abscissa zero and is written (0, y). Problems ask you to find such points at a set distance from a fixed point.
Take the points on the x-axis at distance 10 from (2, −8). Let the point be (x, 0):
10 = \sqrt{(x - 2)^2 + (0 + 8)^2}
Squaring gives (x - 2)^2 = 36, so x = 8 or x = −4. The required points are (8, 0) and (−4, 0).
The equidistant point method
A point equidistant from two given points satisfies PA = PB. Square both sides to remove the square roots. The x^2 and y^2 terms cancel, leaving a linear equation.
Find the point on the y-axis equidistant from (8, 1) and (−2, 7). Let it be (0, y):
\sqrt{64 + (1 - y)^2} = \sqrt{4 + (7 - y)^2}
Squaring and simplifying gives 65 − 2y = 53 − 14y, so 12y = −12 and y = −1. The point is (0, −1).
The collinearity test
Three points A, B and C are collinear if and only if the sum of two of the distances equals the third. Depending on which point lies between the others, the condition is AB + BC = AC, AB + AC = BC, or AC + BC = AB.
For A(2, 3), B(5, 7) and C(8, 11): AB = 5, BC = 5 and AC = 10. Since AB + BC = AC, the points are collinear.
Identifying triangles and quadrilaterals
A triangle is isosceles when two sides are equal. It is isosceles right-angled when two sides are equal and Pythagoras holds, such as AB^2 + AC^2 = BC^2.
For quadrilaterals, the chapter gives four tests:
| Shape | Test |
|---|---|
| Parallelogram | Both pairs of opposite sides equal |
| Rectangle | Both pairs of opposite sides equal, diagonals equal |
| Rhombus | All four sides equal |
| Square | All four sides equal, diagonals equal |
For A(1, 1), B(5, 1) and C(1, 5): AB = 4, AC = 4, and BC = 4\sqrt{2}. Two sides are equal and 4^2 + 4^2 = 32 = BC^2, so the triangle is isosceles right-angled.
Area of a circle from its centre and a point
The radius of a circle is the distance from its centre to any point on the circle. Once you have the radius, area = \pi r^2, using \pi = 3.14.
For a circle with centre (2, 4) passing through (−3, 8):
r = \sqrt{(-3 - 2)^2 + (8 - 4)^2} = \sqrt{25 + 16} = \sqrt{41}
\text{Area} = 3.14 \times 41 = 128.74 \text{ sq units}
Ratio problems
When distances are given in a ratio such as PA : PB = 1 : 2, write 2PA = PB. Square both sides to clear the roots, then solve the quadratic that results.
For P(x, 0) with PA : PB = 1 : 2 from A(5, 2) and B(−2, 4):
2\sqrt{(x - 5)^2 + 4} = \sqrt{(x + 2)^2 + 16}
Squaring gives 3x^2 - 44x + 96 = 0, so x = 12 or x = 8/3. The points are (12, 0) and (8/3, 0).
Watch for these errors when you apply the formula:
- Sign errors: subtracting a negative coordinate changes the sign. Write the full subtraction before simplifying.
- Squaring only one side: when an equation has roots on both sides, square the whole equation.
- Forgetting the final root: the formula ends with a square root, so the answer is the root, not the squared sum.
- Mixing coordinates: x-differences belong with x, y-differences with y.
28.3 CIRCUMCENTRE OF A TRIANGLE
The circumcentre of a triangle is the point equidistant from all three vertices. If P is the circumcentre of triangle ABC, then PA = PB = PC. That common distance is called the circumradius. A circle drawn with P as centre and PA as radius passes through all three vertices.
To find the coordinates of the circumcentre, use two equations. Set PA = PB to get one linear equation, and PA = PC to get a second. Solve the two equations simultaneously.
For a triangle with vertices A(3, 2), B(1, 4) and C(5, 4), let P(x, y) be the circumcentre.
From PA = PB:
\sqrt{(x - 3)^2 + (y - 2)^2} = \sqrt{(x - 1)^2 + (y - 4)^2}
Squaring and simplifying gives −x + y = 1, or y = x + 1. From PA = PC:
\sqrt{(x - 3)^2 + (y - 2)^2} = \sqrt{(x - 5)^2 + (y - 4)^2}
This simplifies to x + y = 7. Solving y = x + 1 with x + y = 7 gives x = 3 and y = 4. The circumcentre is (3, 4), and the circumradius is 2 units. Check that PA = PB = PC = 2. The previous chapter on graphical solutions used similar coordinate methods.
What to practise from Distance Formula Class 9 ICSE
The exercise at the end of this chapter has 28 questions. They cover every idea in sections 28.2 and 28.3, from basic substitution to proofs. Working through all of them builds speed with the Class 9 ICSE distance formula.
- Distance between two pairs of points, including negative, fractional and surd coordinates
- Distance from the origin, such as (−8, 6) and (−5, −12)
- Points on the x-axis and y-axis at a given distance from a fixed point
- Finding an unknown coordinate when a distance is given
- Equidistant point problems on both axes
- Collinearity proofs using the sum of two distances
- Isosceles and isosceles right-angled triangle proofs, including area
- Rectangle, square and rhombus vertex proofs
- Quadrilateral conditions such as AB = CD
- Circumcentre coordinates of a triangle
- Circle centre and radius problems, including chord length with a perpendicular from the centre
- Distances correct to three significant figures
- Ratio-of-distances problems showing a given equation
Distance Formula Class 9 ICSE frequently asked questions
What is the distance formula in class 9 ICSE maths?
The distance between two points (x_1, y_1) and (x_2, y_2) is \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. It comes from Pythagoras’ theorem. The order of subtraction does not matter because (a - b)^2 = (b - a)^2.
How do I find the distance of a point from the origin in Class 9 ICSE maths?
Substitute the origin (0, 0) into the formula. The distance of (x, y) from the origin is \sqrt{x^2 + y^2}. For (−8, 15), that gives \sqrt{64 + 225} = 17 units.
How do I prove three points are collinear using the distance formula in this ICSE chapter?
Calculate all three distances. The points are collinear if and only if the sum of two of the distances equals the third, such as AB + BC = AC. Check each pairing until one holds.
What is the circumcentre of a triangle in class 9 ICSE maths?
The circumcentre is the point equidistant from all three vertices, so PA = PB = PC. Find it by setting PA = PB and PA = PC, simplifying each to a linear equation, then solving the pair simultaneously.
How do I find points on the x-axis at a given distance in the Class 9 ICSE distance formula chapter?
Let the point be (x, 0) and substitute into the formula with the given distance. Square both sides, solve the quadratic, and you get two points. The same method works for the y-axis with (0, y).
Reference: ICSE Class 9 Mathematics textbook material, chapter 28: Distance Formula — CISCE-aligned.
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