Class 10 Mathematics · Chapter 7 NotesCoordinate Geometry
Revise Class 10 Mathematics Coordinate Geometry with clear notes on the distance formula, section formula, midpoint, area of a triangle, and linear equations.
Coordinate Geometry is the branch of mathematics that uses algebra to study geometry. In this chapter, you will learn how to locate points on a plane using a pair of coordinate axes, and how to find distances and division points using coordinates. You will begin by recalling the Cartesian system, where the x-coordinate (abscissa) and y-coordinate (ordinate) describe a point's position. Then you will derive and apply the distance formula to find the distance between two points, including points on the axes and in different quadrants. Next, you will study the section formula, which gives the coordinates of a point dividing a line segment in a given ratio, and its special case, the midpoint formula. You will also see how these ideas connect to linear equations in two variables and how they help solve problems about triangles, quadrilaterals, and real-life situations such as locating a relay tower or arranging desks in a classroom.
What you'll learn
1Recall the Cartesian system and identify coordinates of points on the axes and in quadrants.
2Derive and apply the distance formula to find the distance between two points.
3Use the distance formula to verify collinearity, types of triangles, and types of quadrilaterals.
4Derive and apply the section formula to find coordinates of a point dividing a segment in a given ratio.
5Use the midpoint formula as a special case of the section formula.
6Solve problems involving points of trisection and ratios in which axes divide a segment.
7Relate coordinate geometry to linear equations in two variables and their graphs.
Chapter at a glance
01Cartesian System of Coordinates
02Distance Formula
03Section Formula
04Area of a Triangle
05Linear Equations in Two Variables
Detailed chapter notes
01
Cartesian System of Coordinates
To locate a point on a plane, we use two perpendicular number lines called the x-axis and the y-axis. Their intersection is the origin O(0, 0). The distance of a point from the y-axis is its x-coordinate or abscissa, and the distance from the x-axis is its y-coordinate or ordinate. A point is written as an ordered pair (x, y). Points on the x-axis have the form (x, 0), and points on the y-axis have the form (0, y). The axes divide the plane into four quadrants. The signs of the coordinates tell us the quadrant: (+, +) in the first, (–, +) in the second, (–, –) in the third, and (+, –) in the fourth. This system lets us describe geometric figures using algebra.
OriginO(0, 0)
Point on x-axis(x, 0)
Point on y-axis(0, y)
Quadrant signsI (+, +), II (–, +), III (–, –), IV (+, –)
02
Distance Formula
The distance between two points P(x₁, y₁) and Q(x₂, y₂) is found using the distance formula: PQ = √[(x₂ – x₁)² + (y₂ – y₁)²]. This formula comes from the Pythagoras theorem. If the points lie on the x-axis, the distance is the difference of their x-coordinates; similarly for the y-axis. The distance of a point P(x, y) from the origin is √(x² + y²). The distance is always non-negative, so we take the positive square root. Using this formula, we can check whether three points are collinear (if the sum of two distances equals the third), identify types of triangles (equilateral, isosceles, right-angled), and verify properties of quadrilaterals such as squares and rhombuses.
Distance formulaPQ = √[(x₂ – x₁)² + (y₂ – y₁)²]
Distance from originOP = √(x² + y²)
CollinearityAB + BC = AC
03
Section Formula
The section formula gives the coordinates of a point P(x, y) that divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂. The coordinates are: x = (m₁x₂ + m₂x₁)/(m₁ + m₂) and y = (m₁y₂ + m₂y₁)/(m₁ + m₂). This formula is derived using similar triangles. If the ratio is k : 1, the coordinates become ((kx₂ + x₁)/(k+1), (ky₂ + y₁)/(k+1)). The section formula is useful for finding points of trisection, the ratio in which an axis divides a segment, and for solving problems involving parallelograms and circles.
Ratio k1: P = ((kx₂ + x₁)/(k+1), (ky₂ + y₁)/(k+1))
04
Midpoint Formula
The midpoint of a line segment divides it in the ratio 1 : 1. Substituting m₁ = m₂ = 1 in the section formula gives the midpoint formula: the coordinates of the midpoint of the segment joining A(x₁, y₁) and B(x₂, y₂) are ((x₁ + x₂)/2, (y₁ + y₂)/2). This is a special case of the section formula and is very useful in problems involving diagonals of parallelograms, circles, and finding points that divide a segment into equal parts.
Midpoint formulaM = ((x₁ + x₂)/2, (y₁ + y₂)/2)
05
Area of a Triangle
Although the chapter summary does not explicitly list the area formula, the chapter includes problems that require finding areas of triangles and quadrilaterals using coordinates. The area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) can be found using the formula: Area = ½ |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|. This formula is derived from the distance formula and the concept of coordinates. It is useful for finding areas of polygons by dividing them into triangles. The absolute value ensures the area is positive.
Area of triangle½ |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
06
Linear Equations in Two Variables
A linear equation in two variables, such as ax + by + c = 0 (where a and b are not both zero), represents a straight line when graphed on the coordinate plane. Every point on the line satisfies the equation, and every solution of the equation is a point on the line. This connects algebra and geometry: we can solve problems by graphing lines and finding intersection points, or by using algebraic methods. In coordinate geometry, linear equations help describe loci, such as the perpendicular bisector of a segment, which is the set of points equidistant from the endpoints. For example, the relation x – y = 2 represents the perpendicular bisector of the segment joining (7, 1) and (3, 5).
General formax + by + c = 0
Graph is a straight line
Points on the line satisfy the equation
Want the complete chapter resources?Topic notes, quizzes and flashcards for Coordinate Geometry.
Q1. Define the terms abscissa and ordinate. What are the coordinates of a point on the x-axis and on the y-axis?
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Model answer
The abscissa is the distance of a point from the y-axis, which is its x-coordinate. The ordinate is the distance of a point from the x-axis, which is its y-coordinate. A point on the x-axis has coordinates (x, 0), and a point on the y-axis has coordinates (0, y).
Sample question3 marks
Q2. Find the distance between the points (2, 3) and (4, 1).
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Model answer
Using the distance formula, distance = √[(4-2)² + (1-3)²] = √[2² + (-2)²] = √(4+4) = √8 = 2√2 units.
Sample question3 marks
Q3. Find the coordinates of the point which divides the line segment joining the points (4, -3) and (8, 5) in the ratio 3 : 1 internally.
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Model answer
Using the section formula, the coordinates of the point P(x, y) are given by x = (3*8 + 1*4)/(3+1) = (24+4)/4 = 7, y = (3*5 + 1*(-3))/(3+1) = (15-3)/4 = 3. Therefore, the required point is (7, 3).
Sample question3 marks
Q4. Find the area of the triangle whose vertices are (1, 2), (3, 4), and (5, 6).
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Model answer
Using the area formula: Area = 1/2 | x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |. Substituting (1,2), (3,4), (5,6): = 1/2 | 1(4-6) + 3(6-2) + 5(2-4) | = 1/2 | -2 + 12 - 10 | = 1/2 |0| = 0. Hence, the area is 0, meaning the points are collinear.
Sample question3 marks
Q5. Find the distance between the points (2, 3) and (4, 1) using the distance formula.
What is the distance formula in coordinate geometry?
The distance formula gives the distance between two points P(x₁, y₁) and Q(x₂, y₂) as PQ = √[(x₂ – x₁)² + (y₂ – y₁)²]. It is derived from the Pythagoras theorem and is used to find lengths of segments, check collinearity, and classify triangles and quadrilaterals.
What is the section formula?
The section formula gives the coordinates of a point P(x, y) that divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂. The coordinates are x = (m₁x₂ + m₂x₁)/(m₁ + m₂) and y = (m₁y₂ + m₂y₁)/(m₁ + m₂).
How do you find the midpoint of a line segment?
The midpoint of the segment joining A(x₁, y₁) and B(x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2). This is a special case of the section formula when the ratio is 1 : 1.
How to check if three points are collinear?
Three points are collinear if the sum of the distances between two pairs of points equals the distance between the remaining pair. For example, points A, B, C are collinear if AB + BC = AC (or any other combination that sums to the third distance).
What is the area of a triangle using coordinates?
The area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is ½ |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|. The absolute value ensures the area is positive.
How is coordinate geometry related to linear equations?
A linear equation in two variables, ax + by + c = 0, represents a straight line on the coordinate plane. Every solution of the equation is a point on the line, and every point on the line satisfies the equation. This connects algebra and geometry.