Coordinate Geometry — Class 10 Maths (Samacheer Kalvi)
TN State Board (Samacheer Kalvi) Class 10 Mathematics, Chapter 5. Points, areas, slopes and the equations of lines.
1. About this chapter
This chapter covers the area of a triangle and a quadrilateral, collinearity, slope, and the equation of a straight line in its various forms.
2. Area and collinearity
- Area of a triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃): ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.
- Collinear points: three points are collinear if the area of the triangle = 0.
- Area of a quadrilateral is found by splitting it into triangles (or the shoelace formula).
3. Slope of a line
- Slope m = (y₂ − y₁)/(x₂ − x₁) = tan θ, where θ is the angle of inclination.
- Parallel lines: equal slopes (m₁ = m₂).
- Perpendicular lines: m₁ · m₂ = −1.
4. Equations of a straight line
| Form | Equation |
|---|---|
| Slope–intercept | y = m x + c |
| Point–slope | y − y₁ = m(x − x₁) |
| Two–point | (y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁) |
| Intercept | x/a + y/b = 1 |
| General | a x + b y + c = 0 |
5. Worked examples
Example 1. Find the area of the triangle with vertices (1, 1), (2, 3), (4, 5). = ½|1(3−5) + 2(5−1) + 4(1−3)| = ½|−2 + 8 − 8| = ½|−2| = 1 sq unit.
Example 2. Find the slope of the line joining (2, 3) and (5, 9). m = (9 − 3)/(5 − 2) = 6/3 = 2.
Example 3. Find the equation of the line with slope 2 passing through (1, 3). y − 3 = 2(x − 1) → y = 2x + 1.
6. Common mistakes
- Mistake: Dropping the modulus in the area formula. Fix: Area is always positive — use |…|.
- Mistake: Saying parallel lines have m₁ m₂ = −1. Fix: Parallel → equal slopes; perpendicular → m₁ m₂ = −1.
- Mistake: Mixing up the line forms. Fix: Choose the form that matches the given data.
7. Practice (book-back style)
- Write the formula for the area of a triangle in coordinate geometry.
- Find the slope of the line joining (1, 2) and (4, 8).
- Show that (1, 1), (2, 2), (3, 3) are collinear.
- Find the equation of the line with slope 3 and y-intercept 2.
- State the condition for two lines to be perpendicular.
8. Answer key
- ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.
- m = (8 − 2)/(4 − 1) = 6/3 = 2.
- Area = ½|1(2−3) + 2(3−1) + 3(1−2)| = ½|−1 + 4 − 3| = 0 → collinear.
- y = 3x + 2.
- The product of their slopes m₁ m₂ = −1.
9. Quick revision
- Chapter 5 · area, collinearity, slope, line equations.
- Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|; collinear ⇒ area 0.
- Slope m = (y₂−y₁)/(x₂−x₁) = tan θ.
- Parallel: m₁ = m₂; perpendicular: m₁ m₂ = −1.
- Line forms: y = mx + c; y − y₁ = m(x − x₁); x/a + y/b = 1; ax + by + c = 0.
