For the 2025–2026 academic year, the most important topics in Class 10 Maths for

Coordinate Geometry involve calculating distance, using the section formula, and understanding the conditions for collinear points. The topic of the area of a triangle has been deleted from the syllabus. 

Core topics in coordinate geometry 

Distance formula 

  • You must be able to find the distance between any two points (x1,y1)open paren x sub 1 comma y sub 1 close paren(π‘₯1,𝑦1) and (x2,y2)open paren x sub 2 comma y sub 2 close paren(π‘₯2,𝑦2) using the formula:
    D=(x2βˆ’x1)2+(y2βˆ’y1)2cap D equals the square root of open paren x sub 2 minus x sub 1 close paren squared plus open paren y sub 2 minus y sub 1 close paren squared end-root𝐷=(π‘₯2βˆ’π‘₯1)2+(𝑦2βˆ’π‘¦1)2√.
  • Special cases: Practice finding the distance of a point from the origin, or questions where you have to find an unknown coordinate by using the fact that two points are equidistant.
  • Applications: Use the distance formula to solve problems related to geometric figures, such as proving if a figure is a square, rhombus, or isosceles triangle by checking the lengths of its sides.Β 

Section formula 

  • This formula is used to find the coordinates of a point that divides a line segment in a given ratio. If a point P(x,y)cap P open paren x comma y close paren𝑃(π‘₯,𝑦) divides the line segment joining points A(x1,y1)cap A open paren x sub 1 comma y sub 1 close paren𝐴(π‘₯1,𝑦1) and B(x2,y2)cap B open paren x sub 2 comma y sub 2 close paren𝐡(π‘₯2,𝑦2) internally in the ratio m∢nm colon nπ‘šβˆΆπ‘›, the coordinates of Pcap P𝑃 are:
    P(x,y)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)cap P open paren x comma y close paren equals open paren the fraction with numerator m sub 1 x sub 2 plus m sub 2 x sub 1 and denominator m sub 1 plus m sub 2 end-fraction comma the fraction with numerator m sub 1 y sub 2 plus m sub 2 y sub 1 and denominator m sub 1 plus m sub 2 end-fraction close paren𝑃(π‘₯,𝑦)=π‘š1π‘₯2+π‘š2π‘₯1π‘š1+π‘š2,π‘š1𝑦2+π‘š2𝑦1π‘š1+π‘š2.
  • Midpoint formula: This is a special case of the section formula where the ratio is 1:1.
  • Finding the ratio: You should be able to solve problems where you have to find the ratio in which a given point divides a line segment.
  • Points of trisection: Trisection points are a common application of the section formula, where the ratio is 1:2 and 2:1.Β 

Collinearity of three points 

  • Three points are collinear if they lie on the same straight line.
  • Using the distance formula: To test for collinearity, show that the distance between the two extreme points is equal to the sum of the distances of the intermediate points.
  • Using the section formula: If the section ratio is the same for both the x- and y-coordinates, the points are collinear.Β 

Deleted topic for 2025–2026 

  • Area of a triangle: The formula for finding the area of a triangle when the coordinates of its vertices are given has been removed from the syllabus. While related concepts like testing for collinearity are still important, direct questions using the area formula are not expected.Β 

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Last Update: 20/10/2025