Chapter 7: Coordinate Geometry

Overview

Coordinate geometry combines algebraic techniques with geometric concepts to study properties of geometric figures using coordinate systems. This chapter explores fundamental topics including division of line segments, parallel and perpendicular lines, area calculations of polygons, and locus equations. Mastery of coordinate geometry is essential for solving geometric problems using algebraic methods and forms the foundation for advanced topics in calculus and vector analysis.

Learning Objectives

After completing this chapter, you will be able to:

• Divide line segments in given ratios using the section formula
• Apply conditions for parallel and perpendicular lines
• Calculate areas of polygons using coordinate methods
• Find equations of loci for various geometric conditions
• Use coordinate geometry to solve geometric problems

Key Concepts

7.1 Division of a Line Segment

Section Formula

The section formula finds the coordinates of a point P(x, y) that divides the line segment joining A($x_1$, $y_1$) and B($x_2$, $y_2$) in the ratio m:n.

$$P(x, y) = \left( \frac{nx_1 + mx_2}{m+n}, \frac{ny_1 + my_2}{m+n} \right)$$

Special Cases:

• Midpoint Formula: When m = n = 1

$$M(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$

• Internal Division: Point between A and B
• External Division: Point outside AB (using negative ratios)

Distance Formula

The distance between points A($x_1$, $y_1$) and B($x_2$, $y_2$) is:

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

7.2 Parallel Lines and Perpendicular Lines

Gradient (Slope)

The gradient of a line joining points A($x_1$, $y_1$) and B($x_2$, $y_2$) is:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}$$

Conditions for Parallel Lines

Two lines are parallel if they have the same gradient:

$$m_1 = m_2$$

Conditions for Perpendicular Lines

Two lines are perpendicular if the product of their gradients is -1:

$$m_1 \times m_2 = -1$$

Equation of Lines

1. Two-point form: Using points ($x_1$, $y_1$) and ($x_2$, $y_2$)

$$\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$$

2. Point-gradient form: Using point ($x_1$, $y_1$) and gradient m

$$y - y_1 = m(x - x_1)$$

3. Gradient-intercept form: $y = mx + c$

7.3 Area of Polygons

Shoelace Formula

For a polygon with vertices ($x_1$, $y_1$), ($x_2$, $y_2$), ..., (xₙ, yₙ) arranged in order:

$$\text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n} (x_i y_{i+1}) - \sum_{i=1}^{n} (y_i x_{i+1}) \right|$$

Where $x_{n+1} = x_1$ and $y_{n+1} = y_1$.

Triangle Area

For triangle with vertices A($x_1$, $y_1$), B($x_2$, $y_2$), C($x_3$, $y_3$):

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

7.4 Equation of a Locus

Definition of Locus

A locus is a set of points that satisfy one or more geometric conditions.

Common Loci

1. Distance from a fixed point: Circle Distance from A($x_1$, $y_1$) is constant k:

   $$\sqrt{(x - x_1)^2 + (y - y_1)^2} = k$$

2. Distance from two fixed points: Ellipse or perpendicular bisector Distance from A equals distance from B:

   $$\sqrt{(x - x_1)^2 + (y - y_1)^2} = \sqrt{(x - x_2)^2 + (y - y_2)^2}$$

3. Constant ratio of distances: Circle (Apollonius circle) PA/PB = k (constant ratio)

Important Formulas and Methods

Key Formulas Summary

Problem-Solving Strategies

Coordinate Geometry Problems:

1. Sketch the figure with given points
2. Calculate distances and gradients
3. Apply appropriate formulas
4. Check geometric properties
5. Verify results

Locus Problems:

1. Let P(x, y) be a variable point
2. Apply the given geometric condition
3. Form an equation relating x and y
4. Simplify to standard form

Solved Examples

Example 1: Section Formula and Midpoint

Find the coordinates of the point that divides the line segment joining A(2, 3) and B(8, 9) in the ratio 3:2.

Solution:

Using section formula with m:n = 3:2 P(x, y) = ((2×2 + 3×8)/(3+2), (2×3 + 3×9)/(3+2)) = ((4 + 24)/5, (6 + 27)/5) = (28/5, 33/5) = (5.6, 6.6)

Midpoint verification: M = ((2+8)/2, (3+9)/2) = (5, 6)

Example 2: Parallel and Perpendicular Lines

Given line $L_1$: 2x + 3y = 6 a) Find equation of line parallel to $L_1$ passing through (1, 4) b) Find equation of line perpendicular to $L_1$ passing through (1, 4)

Solution:

a) First, find gradient of $L_1$: 2x + 3y = 6 ⇒ 3y = -2x + 6 ⇒ y = (-2/3)x + 2 Gradient $m_1$ = -2/3

Parallel line has same gradient m = -2/3 Using point-slope form: y - 4 = (-2/3)(x - 1) y = (-2/3)x + 2/3 + 4 y = (-2/3)x + 14/3 Multiply by 3: 3y = -2x + 14 ⇒ 2x + 3y = 14

b) Perpendicular line has gradient $m_2$ where $m_1$ × $m_2$ = -1 (-2/3) × $m_2$ = -1 ⇒ $m_2$ = 3/2

Using point-slope form: y - 4 = (3/2)(x - 1) y = (3/2)x - 3/2 + 4 y = (3/2)x + 5/2 Multiply by 2: 2y = 3x + 5 ⇒ 3x - 2y + 5 = 0

Example 3: Area of Triangle

Find the area of triangle with vertices A(1, 2), B(4, 6), C(7, 2).

Solution:

Using shoelace formula: Area = ½|$x_1$($y_2$ - $y_3$) + $x_2$($y_3$ - $y_1$) + $x_3$($y_1$ - $y_2$)| = ½|1(6 - 2) + 4(2 - 2) + 7(2 - 6)| = ½|1(4) + 4(0) + 7(-4)| = ½|4 + 0 - 28| = ½| -24 | = ½ × 24 = 12 square units

Example 4: Locus of Points

Find the equation of the locus of a point P(x, y) that is equidistant from A(2, 3) and B(6, 1).

Solution:

Distance PA = Distance PB √[(x - 2)² + (y - 3)²] = √[(x - 6)² + (y - 1)²]

Square both sides: (x - 2)² + (y - 3)² = (x - 6)² + (y - 1)²

Expand: $x^2$ - 4x + 4 + $y^2$ - 6y + 9 = $x^2$ - 12x + 36 + $y^2$ - 2y + 1

Simplify: -4x - 6y + 13 = -12x - 2y + 37

Bring all terms to one side: 8x - 4y - 24 = 0 Divide by 4: 2x - y - 6 = 0 Or y = 2x - 6

This is the perpendicular bisector of AB.

Example 5: Complex Problem

Find the coordinates of the vertices of a triangle with area 24 square units, given two vertices A(1, 2) and B(5, 6), and the third vertex C on the line y = 3.

Solution:

Let C be (x, 3) since it lies on y = 3.

Using area formula: Area = ½|$x_1$($y_2$ - $y_3$) + $x_2$($y_3$ - $y_1$) + $x_3$($y_1$ - $y_2$)| = 24 ½|1(6 - 3) + 5(3 - 2) + x(2 - 6)| = 24 ½|1(3) + 5(1) + x(-4)| = 24 ½|3 + 5 - 4x| = 24 ½|8 - 4x| = 24 |8 - 4x| = 48

This gives two cases:

1. 8 - 4x = 48 ⇒ -4x = 40 ⇒ x = -10
2. 8 - 4x = -48 ⇒ -4x = -56 ⇒ x = 14

So C is either (-10, 3) or (14, 3).

Mathematical Derivations

Derivation of Section Formula

Let A($x_1$, $y_1$) and B($x_2$, $y_2$) divide the line segment in ratio m:n. The coordinates of P are weighted averages: x = (n$x_1$ + m$x_2$)/(m + n) y = (n$y_1$ + m$y_2$)/(m + n)

This follows from the concept of weighted means where the weights are the ratios.

Derivation of Perpendicular Line Condition

If line 1 has slope $m_1$ and line 2 has slope $m_2$, they are perpendicular if the angle between them is 90°.

Using the tangent of angle between lines: tan θ = |($m_2$ - $m_1$)/(1 + $m_1m_2$)|

For θ = 90°, tan θ is undefined, so 1 + $m_1m_2$ = 0 Therefore: $m_1m_2$ = -1

Derivation of Shoelace Formula

For a polygon with vertices ($x_1$, $y_1$), ($x_2$, $y_2$), ..., (xₙ, yₙ):

The area can be calculated by dividing into triangles or using the surveyor's formula: Area = ½|Σ(x_i y_{i+1}) - Σ(y_i x_{i+1})|

This formula works because it calculates the sum of trapezoidal areas between consecutive points.

Real-World Applications

1. Computer Graphics

Coordinate geometry is used to:

• Position objects in 2D/3D space
• Transform objects (translation, rotation, scaling)
• Detect collisions and intersections
• Create smooth curves and surfaces

2. Navigation Systems

GPS and navigation use:

• Coordinate systems to locate positions
• Distance calculations for route planning
• Intersection calculations for determining paths
• Area calculations for coverage planning

3. Architecture and Engineering

Applications include:

• Precise positioning of structural elements
• Area calculations for material estimation
• Distance and angle calculations for structural analysis
• Computer-aided design (CAD) systems

4. Physics and Motion

Coordinate geometry helps in:

• Analyzing projectile motion
• Calculating work done in force fields
• Determining equilibrium positions
• Modeling wave interference patterns

Complex Problem-Solving Techniques

Problem: Find the equation of the line passing through the intersection of 2x + 3y = 7 and x - y = 1, and perpendicular to 3x + 4y = 8.

Solution:

First, find intersection point: From x - y = 1 ⇒ x = y + 1 Substitute into 2x + 3y = 7: 2(y + 1) + 3y = 7 ⇒ 2y + 2 + 3y = 7 ⇒ 5y = 5 ⇒ y = 1 Then x = 1 + 1 = 2 Intersection point: (2, 1)

Now, find slope of given line: 3x + 4y = 8 ⇒ 4y = -3x + 8 ⇒ y = (-3/4)x + 2 Slope = -3/4

Perpendicular slope: m = 4/3 (negative reciprocal)

Equation: y - 1 = (4/3)(x - 2) 3y - 3 = 4x - 8 4x - 3y - 5 = 0

Problem: Find the area of quadrilateral with vertices A(1,1), B(4,2), C(5,5), D(2,6).

Solution:

Using shoelace formula for polygon: List vertices in order (clockwise or counterclockwise): A(1,1), B(4,2), C(5,5), D(2,6), A(1,1)

Area = ½|(1×2 + 4×5 + 5×6 + 2×1) - (1×4 + 2×5 + 5×2 + 6×1)| = ½|(2 + 20 + 30 + 2) - (4 + 10 + 10 + 6)| = ½|54 - 30| = ½ × 24 = 12 square units

Problem: Find the locus of a point P such that the distance from P to A(1,3) is twice the distance from P to B(4,1).

Solution:

PA = 2 × PB √[(x-1)² + (y-3)²] = 2√[(x-4)² + (y-1)²]

Square both sides: (x-1)² + (y-3)² = 4[(x-4)² + (y-1)²]

Expand: $x^2$ - 2x + 1 + $y^2$ - 6y + 9 = 4[$x^2$ - 8x + 16 + $y^2$ - 2y + 1] $x^2$ - 2x + $y^2$ - 6y + 10 = 4$x^2$ - 32x + 4$y^2$ - 8y + 68

Bring all terms to left: -3$x^2$ + 30x - 3$y^2$ + 2y - 58 = 0 Multiply by -1: 3$x^2$ - 30x + 3$y^2$ - 2y + 58 = 0 Divide by 3: $x^2$ - 10x + $y^2$ - (2/3)y + 58/3 = 0

Complete the square: $x^2$ - 10x + 25 + $y^2$ - (2/3)y + (1/9) = -58/3 + 25 + 1/9 (x - 5)² + (y - 1/3)² = (-174/9) + 225/9 + 1/9 = 52/9

This is a circle with center (5, 1/3) and radius √(52/9) = (2√13)/3

Summary Points

• Coordinate geometry combines algebra and geometry to solve geometric problems
• The section formula divides line segments in given ratios
• Parallel lines have equal slopes, perpendicular lines have slopes whose product is -1
• The shoelace formula calculates polygon areas from coordinates
• Locus problems find equations describing sets of points with specific geometric properties
• These techniques have wide applications in science, engineering, and technology

Common Mistakes to Avoid

1. Ratio confusion - In section formula, m:n means m parts to n parts, not m/n
2. Perpendicular slope errors - Remember it's negative reciprocal, not just reciprocal
3. Area sign errors - Take absolute value of shoelace formula result
4. Locus setup errors - Start with variable point P(x,y) and apply conditions correctly
5. Order mistakes - In shoelace formula, vertices must be ordered properly

SPM Exam Tips

Exam Strategies

1. Master basic formulas - Distance, midpoint, section, area formulas
2. Practice gradient calculations - Essential for parallel/perpendicular lines
3. Draw diagrams - Visual representation helps coordinate geometry problems
4. Show all steps - Working is crucial for partial marks
5. Check units - Include units for area calculations

Key Exam Topics

• Section formula and midpoint (20% of questions)
• Parallel and perpendicular lines (25% of questions)
• Area calculations (25% of questions)
• Locus equations (20% of questions)
• Complex applications (10% of questions)

Time Management Tips

• Basic coordinate problems: 3-4 minutes
• Gradient and line problems: 4-5 minutes
• Area calculations: 4-5 minutes
• Locus problems: 6-8 minutes
• Complex applications: 8-10 minutes

Practice Problems

Level 1: Basic Coordinate Geometry

1. Find the coordinates of the point dividing A(3,4) and B(9,12) in the ratio 2:3.

2. Find the midpoint of the line segment joining P(-2,5) and Q(6,-1).

3. Calculate the distance between points A(1,2) and B(5,8).

Level 2: Lines and Slopes

4. Find the equation of the line parallel to 3x - 2y = 6 passing through (2,5).

5. Find the equation of the line perpendicular to 4x + 3y = 12 passing through (-1,2).

6. Find the slope of the line passing through A(3,-2) and B(-1,4).

Level 3: Area Calculations

7. Find the area of triangle with vertices A(0,0), B(4,0), C(0,3).

8. Calculate the area of quadrilateral with vertices A(1,1), B(5,2), C(6,6), D(2,5).

9. Use shoelace formula to find the area of pentagon with vertices (1,2), (3,5), (6,4), (5,1), (2,0).

Level 4: Locus Problems

10. Find the equation of the locus of points equidistant from A(2,3) and B(6,1).

11. Find the locus of a point P such that PA = 3, where A is (1,1).

12. Find the equation of the locus of points P where the distance to (0,2) is equal to the distance to the x-axis.

Level 5: Applications

13. Triangle Area: A triangle has vertices A(2,3), B(5,7), and C(x,y). The area is 15 square units. If C lies on the line y = 2x + 1, find x.

14. Rectangle: Find the coordinates of the fourth vertex D of rectangle ABCD if A(1,2), B(4,6), C(7,3).

15. Locus Application: Find the locus of a point P such that the sum of distances to A(0,0) and B(4,0) is 6.

Did You Know? 📚

Coordinate geometry was developed by René Descartes in the 17th century, revolutionizing mathematics by allowing geometric problems to be solved algebraically. The famous Cartesian coordinate system is named after him, and this fusion of algebra and geometry laid the foundation for modern calculus and mathematical physics. Descartes' work "La Géométrie" published in 1637 introduced this powerful analytical method.

Quick Reference Guide

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Coordinate geometry provides a powerful bridge between algebraic and geometric thinking, enabling precise mathematical analysis of spatial relationships. Mastering these techniques will serve you well in advanced mathematics and its applications in science and engineering.