CBSE Class 10 Mathematics Coordinate Geometry Notes
About This Chapter
Chapter 7 - Coordinate Geometry is a fascinating chapter that bridges algebra and geometry. By placing geometric figures on a coordinate plane, we gain the power to describe shapes, distances, and positions using numbers and equations. This transforms many visual problems into straightforward calculations.
Coordinate geometry is used in navigation systems (GPS), computer graphics, engineering design, map-making, and data visualisation. Every time a location is described using latitude and longitude, or a game character moves across a screen, coordinate geometry is at work. It is the foundation of analytic geometry studied in higher classes.
This chapter typically carries 6 to 8 marks in the CBSE board examination. Questions range from direct formula-based problems in 1-mark MCQs to multi-step coordinate problems in 3-mark and 5-mark questions. The Distance Formula, Section Formula, and Area of Triangle formula are the three core tools you must master.
What You Will Learn:
• How to find the distance between two points on a coordinate plane using the Distance Formula
• How to find the coordinates of a point dividing a line segment in a given ratio using the Section Formula
• Finding the midpoint of a line segment as a special case of the Section Formula
• Calculating the area of a triangle when the coordinates of its three vertices are known
• Using coordinate geometry to check collinearity, find centroids, and solve real-life geometry problems
These notes include clear concept explanations, step-by-step solved examples, a complete formula summary table, common mistake warnings, and practice questions organised by marks as per the CBSE pattern.
A detailed and printable PDF version of these notes is attached below for your convenience. Download, save, and use it for your revision!
1. Introduction and Basic Definitions
Coordinate geometry (also called Cartesian geometry or analytic geometry) is the study of geometry using a coordinate system. Points, lines, and shapes are represented using numerical coordinates, making it possible to apply algebraic methods to geometric problems.
1.1 The Coordinate Plane
The coordinate plane (also called the Cartesian plane) is formed by two perpendicular number lines: the horizontal x-axis and the vertical y-axis. They intersect at a point called the origin O, with coordinates (0, 0).
Quadrant | x value | y value | Example Point |
Quadrant I | Positive (+) | Positive (+) | (3, 5) |
Quadrant II | Negative (-) | Positive (+) | (-2, 4) |
Quadrant III | Negative (-) | Negative (-) | (-3, -1) |
Quadrant IV | Positive (+) | Negative (-) | (4, -6) |
1.2 Coordinates of a Point
Every point in the plane is described by an ordered pair (x, y), where x is the horizontal distance from the origin (called the abscissa) and y is the vertical distance from the origin (called the ordinate).
>> Key Reminder The order in an ordered pair matters. The point (3, 5) is NOT the same as (5, 3). The first value is always the x-coordinate (abscissa) and the second is always the y-coordinate (ordinate). |
2. Key Concepts and Components
2.1 Distance Formula
The Distance Formula gives the length of the line segment joining any two points in a coordinate plane. It is derived using the Pythagoras theorem applied to the right triangle formed by the two points and their horizontal/vertical projections.
>> Derivation Idea For points A(x1, y1) and B(x2, y2), draw a horizontal line from A and a vertical line from B. They meet at C(x2, y1). Then AC = |x2 - x1| and BC = |y2 - y1|. By Pythagoras: AB^2 = AC^2 + BC^2, giving the Distance Formula. |
2.2 Section Formula
The Section Formula gives the coordinates of a point P that divides the line segment joining two points A(x1, y1) and B(x2, y2) in the ratio m:n internally.
>> Key Insight Internal division means the point P lies BETWEEN A and B on the segment. External division (not in Class 10 syllabus) means P lies outside the segment on its extension. |
2.3 Midpoint Formula
The Midpoint Formula is a special case of the Section Formula where the ratio is 1:1, meaning the point divides the segment into two equal halves.
2.4 Area of a Triangle
When the vertices of a triangle are known as coordinates, the area can be calculated directly without needing the base and height. This formula uses the cross-multiplication arrangement of coordinates.
>> Collinearity Check If three points are collinear (lie on the same straight line), they do not form a triangle. In that case, the area formula gives 0. This is a very useful way to check if three given points are collinear. |
3. Core Formulas with Derivations
3.1 Distance Formula
Distance Formula Distance between A(x1, y1) and B(x2, y2):
AB = sqrt[ (x2 - x1)^2 + (y2 - y1)^2 ]
Special case - Distance from Origin O(0,0) to P(x, y): OP = sqrt[ x^2 + y^2 ] |
3.2 Section Formula (Internal Division)
Section Formula Point P dividing AB in ratio m:n internally: A(x1, y1), B(x2, y2), ratio m:n
P = ( (m*x2 + n*x1) / (m+n) , (m*y2 + n*y1) / (m+n) )
Memory tip: numerator uses m with the SECOND point (B) and n with the FIRST point (A). |
3.3 Midpoint Formula
Midpoint Formula Midpoint M of segment joining A(x1, y1) and B(x2, y2):
M = ( (x1 + x2) / 2 , (y1 + y2) / 2 )
This is Section Formula with m = n = 1. |
3.4 Area of a Triangle
Area of Triangle Formula Vertices: A(x1, y1), B(x2, y2), C(x3, y3)
Area = (1/2) |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
Note: Take the absolute value. Area is always positive. If Area = 0, the three points are collinear. |
3.5 Centroid of a Triangle
Centroid Formula Centroid G of triangle with vertices A(x1,y1), B(x2,y2), C(x3,y3):
G = ( (x1 + x2 + x3) / 3 , (y1 + y2 + y3) / 3 )
The centroid divides each median in the ratio 2:1 from vertex. |
4. Solved Examples
4.1 Distance Formula
Example 4.1 - Find the distance between points A(3, 4) and B(7, 1) Using Distance Formula: AB = sqrt[ (7 - 3)^2 + (1 - 4)^2 ] = sqrt[ (4)^2 + (-3)^2 ] = sqrt[ 16 + 9 ] = sqrt[ 25 ] = 5 units
Answer: AB = 5 units |
Example 4.2 - Check if points A(1, 5), B(2, 3), C(-2, 11) are collinear using distance AB = sqrt[(2-1)^2 + (3-5)^2] = sqrt[1+4] = sqrt[5] BC = sqrt[(-2-2)^2 + (11-3)^2] = sqrt[16+64] = sqrt[80] = 4*sqrt[5] AC = sqrt[(-2-1)^2 + (11-5)^2] = sqrt[9+36] = sqrt[45] = 3*sqrt[5]
Check: AB + AC = sqrt[5] + 3*sqrt[5] = 4*sqrt[5] = BC
Since AB + AC = BC, the points ARE collinear. Answer: A, B, C are collinear. |
Example 4.3 - Find the point on the x-axis equidistant from A(2, -5) and B(-2, 9) Any point on x-axis has y = 0. Let the point be P(x, 0). Given: PA = PB
PA^2 = (x-2)^2 + (0+5)^2 = x^2 - 4x + 4 + 25 = x^2 - 4x + 29 PB^2 = (x+2)^2 + (0-9)^2 = x^2 + 4x + 4 + 81 = x^2 + 4x + 85
Setting PA^2 = PB^2: x^2 - 4x + 29 = x^2 + 4x + 85 -8x = 56 => x = -7
Answer: The point is (-7, 0). |
4.2 Section Formula
Example 4.4 - Find coordinates of point dividing A(-1, 7) and B(4, -3) in ratio 2:3 Using Section Formula with m = 2, n = 3: x = (m*x2 + n*x1) / (m+n) = (2*4 + 3*(-1)) / (2+3) = (8 - 3) / 5 = 5/5 = 1 y = (m*y2 + n*y1) / (m+n) = (2*(-3) + 3*7) / (2+3) = (-6 + 21) / 5 = 15/5 = 3
Answer: The point is (1, 3). |
Example 4.5 - Find the midpoint of A(4, -6) and B(-2, 8) Using Midpoint Formula: M = ( (4 + (-2))/2 , (-6 + 8)/2 ) = ( 2/2 , 2/2 ) = ( 1 , 1 )
Answer: Midpoint = (1, 1) |
4.3 Area of a Triangle
Example 4.6 - Find the area of triangle with vertices A(1, 1), B(3, 7), C(5, 3) Using Area Formula: Area = (1/2) |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| = (1/2) |1(7 - 3) + 3(3 - 1) + 5(1 - 7)| = (1/2) |1(4) + 3(2) + 5(-6)| = (1/2) |4 + 6 - 30| = (1/2) |-20| = (1/2) * 20 = 10 sq units
Answer: Area = 10 sq units |
Example 4.7 - Check if points P(6, 1), Q(1, 3), R(0, -3) are collinear Area = (1/2) |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| = (1/2) |6(3 - (-3)) + 1((-3) - 1) + 0(1 - 3)| = (1/2) |6(6) + 1(-4) + 0| = (1/2) |36 - 4| = (1/2) * 32 = 16 sq units
Area is NOT 0, so P, Q, R are NOT collinear. Answer: The three points do not lie on the same line. |
5. Applications and Special Cases
5.1 Real-Life Applications
• GPS and Navigation: Every location on Earth is defined by a coordinate pair (latitude, longitude).
• Computer Graphics: Pixels on a screen are addressed using (x, y) coordinates; all rendering uses distance and section formulas.
• Engineering and Architecture: Design plans use coordinate grids; section formula helps place structural elements precisely.
• Gaming: Character positions, collision detection, and movement paths all rely on coordinate geometry.
• Urban Planning: Roads, buildings, and utility lines are planned using coordinate maps.
5.2 Finding the Type of Quadrilateral
One powerful application of the distance formula is determining the type of quadrilateral formed by four given points. Calculate all four sides and both diagonals, then compare:
Quadrilateral | Sides | Diagonals |
Square | All 4 sides equal | Both diagonals equal |
Rectangle | Opposite sides equal | Both diagonals equal |
Rhombus | All 4 sides equal | Diagonals NOT equal |
Parallelogram | Opposite sides equal | Diagonals NOT necessarily equal |
5.3 Trisection of a Line Segment
To find the points that divide a line segment into three equal parts (trisection points), use the Section Formula twice: once with ratio 1:2 and once with ratio 2:1.
Example 5.1 - Find the trisection points of A(1, -2) and B(4, 7) Trisection point P1 divides AB in ratio 1:2: P1 = ( (1*4 + 2*1)/(1+2) , (1*7 + 2*(-2))/(1+2) ) = ( (4+2)/3 , (7-4)/3 ) = ( 6/3 , 3/3 ) = (2, 1)
Trisection point P2 divides AB in ratio 2:1: P2 = ( (2*4 + 1*1)/(2+1) , (2*7 + 1*(-2))/(2+1) ) = ( (8+1)/3 , (14-2)/3 ) = ( 9/3 , 12/3 ) = (3, 4)
Answer: Trisection points are (2, 1) and (3, 4). |
5.4 Finding the Ratio of Division
If a point P(x, y) is given and you need to find in what ratio it divides the segment AB, assume the ratio is k:1 and apply the Section Formula. Solve for k using the x-coordinate equation.
>> Ratio Trick Assume ratio = k:1 (not m:n). Then Section Formula gives x = (k*x2 + x1)/(k+1). Solve for k using the given x-coordinate. Always verify with the y-coordinate to confirm. |
6. Formula Summary Table
All key formulas for Coordinate Geometry at a glance:
Formula Name | Formula | Use Case |
Distance Formula | sqrt[(x2-x1)^2 + (y2-y1)^2] | Length of segment / checking distances |
Distance from Origin | sqrt[x^2 + y^2] | Distance of point from O(0,0) |
Section Formula (internal) | ((mx2+nx1)/(m+n) , (my2+ny1)/(m+n)) | Point dividing AB in ratio m:n |
Midpoint Formula | ((x1+x2)/2 , (y1+y2)/2) | Middle point of a segment |
Area of Triangle | (1/2)|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)| | Area using vertices |
Collinearity Check | Area = 0 | Three points on same line |
Centroid | ((x1+x2+x3)/3 , (y1+y2+y3)/3) | Centre of mass of triangle |
Trisection (1:2) | ((x2+2x1)/3 , (y2+2y1)/3) | First trisection point |
Trisection (2:1) | ((2x2+x1)/3 , (2y2+y1)/3) | Second trisection point |
7. Key Theorems and Properties
>> Property 1 - Distance Formula Derivation The Distance Formula AB = sqrt[(x2-x1)^2 + (y2-y1)^2] is derived directly from the Pythagoras theorem. The horizontal distance |x2-x1| and vertical distance |y2-y1| form the two legs of a right triangle, and AB is the hypotenuse. |
>> Property 2 - Section Formula Derivation The Section Formula is derived using similar triangles. If P divides AB in ratio m:n, two similar triangles can be constructed using horizontal and vertical projections, and the ratio of their sides directly gives the Section Formula. |
>> Property 3 - Area Formula Sign Convention The area formula (1/2)|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)| always gives a positive result due to the absolute value. If the vertices are listed anticlockwise, the expression inside is positive; if clockwise, it is negative. The absolute value handles both cases. |
>> Property 4 - Collinearity Three points A, B, C are collinear if and only if the area of triangle ABC is zero. Equivalently, the slope between any two pairs of points must be equal. In CBSE, the area method is the most direct approach. |
7.1 Properties of Special Points
Point | Definition | Formula |
Midpoint | Divides segment in 1:1 | ((x1+x2)/2, (y1+y2)/2) |
Centroid | Intersection of medians | ((x1+x2+x3)/3, (y1+y2+y3)/3) |
Origin | Intersection of axes | (0, 0) |
x-axis point | y-coordinate = 0 | (a, 0) for any a |
y-axis point | x-coordinate = 0 | (0, b) for any b |
8. Common Mistakes and Exam Tips
8.1 Frequent Errors to Avoid
1. Forgetting the absolute value in the Area Formula - area is always positive, never negative.
2. Swapping x and y coordinates when reading a point - always read as (x, y), horizontal first.
3. In the Section Formula, mixing up which coordinates belong to the first and second point.
4. Not squaring the differences in the Distance Formula - a common calculation error.
5. Concluding collinearity only from 2 distance checks without a third confirmation.
6. Forgetting to take sqrt at the end of the Distance Formula calculation.
8.2 Exam Strategy Tips
• Write the formula first, then substitute - this earns step marks even if the final answer has a calculation error.
• For quadrilateral type questions, always compute all 4 sides AND both diagonals before concluding.
• When checking collinearity, the Area = 0 method is faster than the slope method in most cases.
• For ratio questions, use k:1 substitution to reduce the number of unknowns to one.
• Label all given points clearly at the start - x1, y1, x2, y2 - before substituting.
• Double-check signs when substituting negative coordinates - a common source of errors.
9. Practice Questions
Questions are organised by marks as in CBSE board examinations. Attempt all sections for complete board exam preparation.
9.1 - 1 Mark Questions (MCQ / Very Short Answer)
1. Find the distance between points A(0, 0) and B(3, 4).
2. What is the midpoint of the segment joining (-4, 2) and (6, 8)?
3. Find the coordinates of the point that divides (1, 3) and (2, 7) in the ratio 3:4 internally.
4. If A(2, 3), B(4, k) and AB = 2*sqrt(5), find k.
5. If the area of triangle with vertices (1, 0), (6, 0), (0, m) is 15 sq units, find m.
6. The point (-3, 5) lies in which quadrant?
9.2 - 3 Mark Questions (Short Answer)
1. Find the point on the y-axis equidistant from the points A(6, 5) and B(-4, 3).
2. If points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram taken in order, find the value of p.
3. Find the ratio in which the point P(3/4, 5/12) divides the line segment joining A(1/2, 3/2) and B(2, -5).
4. Find the area of the quadrilateral whose vertices taken in order are (-4, 2), (-3, -5), (3, -2) and (2, 3).
5. Find the coordinates of the centroid of the triangle whose vertices are (4, -3), (-2, 1) and (2, 5).
9.3 - 5 Mark Questions (Long Answer)
1. The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
2. The vertices of a triangle are A(1, 4), B(-2, 2) and C(3, 2). Find the length of the median through vertex A. Also find the centroid.
3. If the coordinates of the midpoints of the sides of a triangle are (1, 1), (2, -3) and (3, 4), find the coordinates of its vertices.
4. If A(-2, 1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Also find the lengths of its sides.
5. A(4, 2), B(6, 5) and C(1, 4) are the vertices of triangle ABC. D is the midpoint of BC and P is a point on AD such that AP = (1/3) AD. Find the coordinates of P.
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