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Mathematics Class 11 - Coordinate Geometry Notes

Comprehensive study notes for Class 11 - Coordinate Geometry olympiad preparation

Coordinate Geometry

Welcome to the chapter on Coordinate Geometry for Class 11. In this chapter, you will learn about the Cartesian coordinate system, plotting points, distance and section formulas, and the basics of straight lines. By the end of this chapter, you will be able to solve problems involving points, lines, and distances in the coordinate plane.

Key Concepts

  • Coordinate Plane: A plane with two perpendicular axes, X and Y, intersecting at the origin (0, 0).
  • Coordinates: An ordered pair (x, y) that shows the position of a point on the plane.
  • Quadrants: The coordinate plane is divided into four quadrants.

Plotting Points

To plot a point (x, y), move x units along the X-axis and y units along the Y-axis. For example, (3, 2) means 3 units right and 2 units up from the origin.

Distance Formula

The distance between two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) is given by:

Distance = \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

Example: Find the distance between (1, 2) and (4, 6):
Distance = \( \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)

Section Formula

The section formula finds the coordinates of a point dividing the line segment joining \(A(x_1, y_1)\) and \(B(x_2, y_2)\) in the ratio \(m:n\):

\( \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \)

Example: Find the point dividing (2, 3) and (8, 7) in the ratio 1:2:
\( \left( \frac{1 \times 8 + 2 \times 2}{1+2}, \frac{1 \times 7 + 2 \times 3}{1+2} \right) = (4, 13/3) \)

Area of a Triangle

The area of a triangle with vertices \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\) is:

\( \text{Area} = \frac{1}{2} | x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) | \)

Example: For points (1, 2), (4, 6), and (5, 3):
Area = \( \frac{1}{2} | 1(6-3) + 4(3-2) + 5(2-6) | = \frac{1}{2} | 3 + 4 - 20 | = \frac{1}{2} \times 13 = 6.5 \)

Equation of a Straight Line

  • Slope-intercept form: \( y = mx + c \), where m is the slope and c is the y-intercept.
  • Point-slope form: \( y - y_1 = m(x - x_1) \)
  • Two-point form: \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \)

Practice Questions

  1. Plot the points (2, 3), (4, 5), and (6, 1) on a graph paper.
  2. Find the distance between the points (1, 2) and (7, 5).
  3. Find the coordinates of the point dividing the line joining (3, 4) and (9, 10) in the ratio 2:1.
  4. Calculate the area of the triangle with vertices (0, 0), (4, 0), and (4, 3).
  5. Write the equation of the line passing through (2, 3) with slope 2.

Challenge Yourself

  • Show that the points (1, 2), (4, 6), and (7, 10) are collinear.
  • Find the equation of the line passing through (1, 2) and (3, 8).
  • If the area of a triangle with vertices (x, 0), (0, y), and (0, 0) is 6, find the relation between x and y.

Did You Know?

  • Coordinate geometry was developed by RenĂ© Descartes, a French mathematician and philosopher.
  • It helps us connect algebra and geometry using graphs and equations.

Glossary

  • Origin: The point (0, 0) where the X and Y axes meet.
  • Quadrant: One of the four sections of the coordinate plane.
  • Slope: The steepness of a line, given by the ratio of vertical change to horizontal change.

Answers to Practice Questions

  1. (Students should plot the points on graph paper.)
  2. Distance = \( \sqrt{(7-1)^2 + (5-2)^2} = \sqrt{36 + 9} = \sqrt{45} = 6.7 \) (approx)
  3. Coordinates = \( \left( \frac{2 \times 9 + 1 \times 3}{2+1}, \frac{2 \times 10 + 1 \times 4}{2+1} \right) = (7, 8) \)
  4. Area = \( \frac{1}{2} \times 4 \times 3 = 6 \) square units
  5. Equation: \( y - 3 = 2(x - 2) \) or \( y = 2x - 1 \)

Practice plotting points and using formulas to master coordinate geometry!

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