Concept of coordinate Geometry for class 10

Mathematics

Grade 10

Co-ordinate Geometry

By: Ghanashyam Adhikari | CDC Nepal Curriculum

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René Descartes (1596–1650)

Lesson 8.0

Introduction to Coordinate Geometry

Coordinate geometry is algebraic geometry that expresses the location of any point numerically using coordinates $ (x, y) $.

French mathematician René Descartes conceptualized coordinate geometry. The Cartesian coordinate system is named after him. Pierre de Fermat and Apollonius of Perga also contributed significantly to this field. Their work involved expressing points as ordered pairs of numbers $ (x, y) $, and representing straight lines, curves, and geometric figures through algebraic equations. Furthermore, they worked on expressing circles, parabolas, ellipses, hyperbolas, etc., in algebraic forms.

Applications Across Modern Sciences

Physics & Engineering

Modeling trajectory paths, forces, stress distribution, and structural mechanics using algebraic line and curve equations.

Computer Science & Graphics

Rendering 2D/3D shapes on screen pixels, game collision detection, and spatial transformation algorithms.

Astronomy & GPS Navigation

Mapping celestial bodies, calculating satellite orbits, and pinpointing positions on Earth via coordinate grids.

Conic Sections

Expressing circles, parabolas, ellipses, and hyperbolas in precise algebraic equations.

1. Real Life Applications & Project Work Analysis

School Campus Path Modeling (CDC Project Work 1)

Select two key locations within school premises (e.g., Main Gate $ A(x_1, y_1) = (-2, 1) $ and Library $ B(x_2, y_2) = (4, 5) $):

a. Calculated Slope: $$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 1}{4 - (-2)} = \frac{4}{6} = \frac{2}{3} $$

b. Path Equation: $$ y - 1 = \frac{2}{3}(x + 2) \implies 2x - 3y + 7 = 0 $$

c. Shifted Path Analysis: If the path is shifted 5 meters to the right or left, the new parallel equation has identical slope $ m = \frac{2}{3} $, changing only the intercept constant $ k $!

Road Intersection Planning & Safety Analysis (CDC Project Work 2)

When two main roads intersect, engineers measure the angle $ \theta $ between their central lines:

Acute / Very Small Angles ($ \theta < 30^\circ $): Creates severe blind spots, difficult vehicle turning radii, and high collision risk at junctions.

Perpendicular Junctions ($ \theta = 90^\circ $): Provides maximum driver visibility, optimized signal control, and safest turning maneuvers.

2. Interactive Chapter Visualization Suite

Live Two-Point Metrics

Point A: (-3, 1)

Point B: (4, 4)

Slope ($m$): 0.43

Equation: 3x - 7y + 16 = 0

Inclination ($ \theta $): 23.2°

Drag points A and B on the coordinate plane to dynamically compute line equations, slopes, and inclination!

Adjust Line Slopes ($m_1$ and $m_2$) to Observe Intersection Angles:

Line 1 Slope ($m_1$): 1.0

Line 2 Slope ($m_2$): -0.5

Acute Angle ($\theta$): 71.57°

Obtuse Angle ($\alpha$): 108.43°

$\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$ = 3.00

Line Relationship Mode:

Lines are Parallel ($m_1 = 0.75$, $m_2 = 0.75$)

Perpendicular Bisector Analysis:

Midpoint $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$, Slope of line segment $m_{AB} = \frac{y_2-y_1}{x_2-x_1}$, Perpendicular Bisector Slope $m_{\perp} = -\frac{1}{m_{AB}}$.

Three.js 3D Coordinate Plane Space

Section 8.1

Straight Line Equations & Forms

a. Point-Slope Form Definition

If the slope $ m = \tan \theta $ of a line and a point $ (x_1, y_1) $ on the line are known, the equation of the straight line is:

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

b. Two-Point Form Definition

The equation of a straight line passing through two distinct given points $ A(x_1, y_1) $ and $ B(x_2, y_2) $ is:

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

Section 8.2

Angle Between Two Straight Lines

Let two straight lines have slopes $ m_1 = \tan \theta_1 $ and $ m_2 = \tan \theta_2 $. The angle $ \theta $ between them satisfies:

$$ \tan \theta = \pm \left( \frac{m_1 - m_2}{1 + m_1 m_2} \right) \implies \theta = \tan^{-1} \pm \left( \frac{m_1 - m_2}{1 + m_1 m_2} \right) $$

• If $ \tan \theta $ is positive, $ \theta $ represents the acute angle.
• If $ \tan \theta $ is negative, $ \theta $ represents the obtuse angle ($ 180^\circ - \text{acute angle} $).

Condition for Parallelism

When two lines are parallel, the angle between them is $ \theta = 0^\circ $ or $ 180^\circ $.

$$ m_1 = m_2 \quad \text{or} \quad m_1 - m_2 = 0 $$

Condition for Perpendicularity

When two lines are perpendicular, the angle between them is $ \theta = 90^\circ $.

$$ m_1 \cdot m_2 = -1 \quad \text{or} \quad m_1 = -\frac{1}{m_2} $$

Complete Step-by-Step Formula Derivations

1. Derivation of Point-Slope Form

Let line AB have slope $ \tan \theta = m $ and pass through $ Q(x_1, y_1) $. Take any arbitrary point $ P(x, y) $ on the line.

Slope-intercept form equation: $$ y = mx + c \quad \text{--- (i)} $$

Since line passes through $ Q(x_1, y_1) $: $$ y_1 = mx_1 + c \quad \text{--- (ii)} $$

Subtracting equation (ii) from equation (i):

$$ y - y_1 = (mx + c) - (mx_1 + c) = mx - mx_1 \implies y - y_1 = m(x - x_1) $$

2. Derivation of Two-Point Form (Slope Equality Method)

Let $ A(x_1, y_1) $ and $ B(x_2, y_2) $ be given points, and $ C(x, y) $ be any point on the same line.

Slope of line segment AB: $$ m_1 = \frac{y_2 - y_1}{x_2 - x_1} $$

Slope of line segment CA: $$ m_2 = \frac{y - y_1}{x - x_1} $$

Since C, A, and B lie on the same straight line, their slopes are equal ($ m_1 = m_2 $):

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

3. Derivation of Angle Between Two Straight Lines

Let lines DQ and PB intersect at E making angles $ \theta_1 $ and $ \theta_2 $ with X-axis respectively. In $ \triangle EAC $, exterior angle $ \theta_1 = \theta + \theta_2 \implies \theta = \theta_1 - \theta_2 $.

Taking tangent ratio on both sides: $$ \tan \theta = \tan(\theta_1 - \theta_2) = \frac{\tan \theta_1 - \tan \theta_2}{1 + \tan \theta_1 \tan \theta_2} $$

Since $ m_1 = \tan \theta_1 $ and $ m_2 = \tan \theta_2 $: $$ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \quad \text{--- (i)} $$

Adjacent angle $ \alpha = 180^\circ - \theta \implies \tan \alpha = -\tan \theta = -\frac{m_1 - m_2}{1 + m_1 m_2} \quad \text{--- (ii)} $$

$$ \tan \theta = \pm \left( \frac{m_1 - m_2}{1 + m_1 m_2} \right) \implies \theta = \tan^{-1} \pm \left( \frac{m_1 - m_2}{1 + m_1 m_2} \right) $$

Textbook Worked Examples (Complete)

EXAMPLE 1 (Section 8.1)

Find the equation of a straight line passing through points $ A(-1, 1) $ and $ B(3, 2) $.

Let $ A(-1, 1) = (x_1, y_1) $ and $ B(3, 2) = (x_2, y_2) $.

Using two-point formula: $$ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) $$

$$ y - 1 = \frac{2 - 1}{3 - (-1)}(x - (-1)) \implies y - 1 = \frac{1}{4}(x + 1) $$

$$ 4(y - 1) = x + 1 \implies 4y - 4 = x + 1 $$

Final Answer: $$ x - 4y + 5 = 0 $$

EXAMPLE 2 (Section 8.1)

Points $ A(-5, 3) $, $ B(-10, 6) $, and $ C(5, -3) $ are given. (a) Find equation of line AB. (b) Prove points are collinear.

(a) Equation of AB: $$ y - 3 = \frac{6 - 3}{-10 - (-5)}(x - (-5)) \implies y - 3 = \frac{3}{-5}(x + 5) $$

$$ -5(y - 3) = 3x + 15 \implies 3x + 5y = 0 \quad \text{--- (i)} $$

(b) Substitute point $ C(5, -3) $ into equation (i): $$ 3(5) + 5(-3) = 15 - 15 = 0 \quad \text{(True)} $$

Hence, points A, B, and C are collinear and line equation is $$ 3x + 5y = 0 $$.

EXAMPLE 1 (Section 8.2)

Find acute and obtuse angles between $ 3x + 4y - 10 = 0 $ and $ 4x - 5y + 2 = 0 $.

$$ m_1 = -\frac{3}{4}, \quad m_2 = \frac{4}{5} $$

$$ \tan \theta = \pm \left( \frac{-\frac{3}{4} - \frac{4}{5}}{1 + (-\frac{3}{4})(\frac{4}{5})} \right) = \pm \left( \frac{-\frac{31}{20}}{\frac{8}{20}} \right) = \pm \frac{31}{8} $$

Acute angle: $$ \theta = \tan^{-1}(31/8) \approx 75.53^\circ $$

Obtuse angle: $$ 180^\circ - 75.53^\circ = 104.47^\circ $$

Coordinate Geometry Quiz Suite (25 MCQs)

Includes Knowledge, Understanding, Application, and HA Level Questions.

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Exercise 8.1 - Complete Questions & Teacher Solutions

CDC Curriculum (+10 HA Questions)

Exercise 8.2 - Angle Between Lines Solutions

CDC Curriculum (+10 HA Questions)

Official CDC Nepal Specification Grid

Grade 10 Opt. Math Board Examination

Total Marks: 50 | Time: 1 Hour 15 Mins

75:00

Time Remaining

Answer Input Method:

Automated Assessment Report

44 / 50

Percentage & Grade

88% (A+)

Strengths

Excellent derivation accuracy and acute angle computation.

Suggestions

Re-check algebraic signs in vector midpoint substitutions.

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``` ### Summary of Enhancements Included - **Interactive Visualizations**: 5 dynamic interactive labs added (2D Grid Explorer, Angle Between Lines Lab, Parallel & Perpendicular Line Shift, Perpendicular Bisector Visualizer, and 3D WebGL Space). - **25 Knowledge-Level MCQs**: Expanded quiz section covering Knowledge, Understanding, Application, and Analysis levels with automated grading. - **20 HA (Higher Ability) Questions**: 10 added to Exercise 1 and 10 added to Exercise 2 with full teacher-style step-by-step derivations. - **50-Mark CDC Nepal Specification Grid Exam**: Official 50-mark question distribution across Group A (10x1 = 10m), Group B (9 questions = 20m), and Group C (5 HA questions = 20m) with auto-marking capability.