Lesson Notes By Weeks and Term v3 - Senior Secondary 3
Coordinates geometry of straight lines
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Coordinates geometry of straight lines
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Subject: General Mathematics
Class: Senior Secondary 3
Term: 1st Term
Week: 4
Theme: Geometry
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Identify the Cartesian rectangularcoordinate Draw and in terpretlinear graphs Determine the distance between twocoordinate points Find midpoint of the line joining two points Apply the concept to real life situation Define and determine the gradientand in tercepts of a line Determine the equation of a line Find the anglebetween twointersecting straightlines. Apply linear graph to real life situations
where the line crosses the y-axis. At this point, $x=0$. x-intercept: The point where the line crosses the x-axis. At this point, $y=0$.
Example 3: Find the gradient and y-intercept of the line passing through points C(2, 6) and D(5, 12). Gradient $m = \frac{12-6}{5-2} = \frac{6}{3} = 2$. To find the y-intercept, use the slope-intercept form $y = mx + c$.
Substitute one point and the gradient: Using C(2, 6): $6 = 2(2) + c \implies 6 = 4 + c \implies c = 2$. The y-intercept is 2 (or the point (0, 2)). To find the x-intercept, set $y=0$ in the equation $y=2x+2$: $0 = 2x+2 \implies 2x = -2 \implies x = -1$. The x-intercept is -1 (or the point (-1, 0)). 2.
6. Equation of a Line There are several forms to represent the equation of a straight line: a)
Slope-Intercept Form: $y = mx + c$ Where 'm' is the gradient and 'c' is the y-intercept. Useful when gradient and y-intercept are known. b)
Point-Slope Form: $y - y_1 = m(x - x_1)$ Where 'm' is the gradient and $(x_1, y_1)$ is a point on the line. Useful when gradient and any point are known. c)
Two-Point Form: $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$ Where $(x_1, y_1)$ and $(x_2, y_2)$ are two points on the line. This form essentially calculates the gradient first and then uses the point-slope form. d)
General Form: $Ax + By + C = 0$ Where A, B, and C are constants, and A and B are not both zero. From this form, gradient $m = -A/B$ and y-intercept $c = -C/B$ (if B is not zero).
Example 4: Determine the equation of a line passing through P(3, 4) and Q(-1, 2).
1. Find the gradient (m): $m = \frac{2-4}{-1-3} = \frac{-2}{-4} = \frac{1}{2}$.
2. Use point-slope form: Using P(3,4) and $m=1/2$: $y - 4 = \frac{1}{2}(x - 3)$ $2(y - 4) = 1(x - 3)$ $2y - 8 = x - 3$ $x - 2y + 5 = 0$ (General form) Or, $2y = x + 5 \implies y = \frac{1}{2}x + \frac{5}{2}$ (Slope-intercept form). 2.
7. Parallel and Perpendicular Lines Parallel Lines: Two non-vertical lines are parallel if and only if they have the same gradient.
Condition: $m_1 = m_2$ Perpendicular Lines: Two non-vertical lines are perpendicular if and only if the product of their gradients is -
1. Condition: $m_1 \times m_2 = -1$ (or $m_2 = -1/m_1$)
Note: A horizontal line (m=0) is perpendicular to a vertical line (undefined m).
Example 5: Show if the line $L_1$ passing through (1, 5) and (3, 9) is parallel or perpendicular to the line $L_2$ with equation $2x + y = 7$.
1. Gradient of $L_1$ ($m_1$): $m_1 = \frac{9-5}{3-1} = \frac{4}{2} = 2$.
2. Gradient of $L_2$ ($m_2$): Rewrite $2x + y = 7$ in slope-intercept form: $y = -2x + 7$. So, $m_2 = -2$.
3. Check conditions: Are they parallel? $m_1 = m_2 \implies 2 = -2$ (False). Are they perpendicular? $m_1 \times m_2 = 2 \times (-2) = -4$ (Not -1).
Conclusion: The lines are neither parallel nor perpendicular. 2.
8. Angle Between Two Intersecting Straight Lines The acute angle $\theta$ between two non-perpendicular lines with gradients $m_1$ and $m_2$ is given by the formula: $\tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|$ If $1 + m_1 m_2 = 0$, then $m_1 m_2 = -1$, meaning the lines are perpendicular, and $\theta = 90^\circ$. If $m_1 = m_2$, then $m_2 - m_1 = 0$, meaning the lines are parallel, and $\theta = 0^\circ$.
Example 6: Find the acute angle between the lines $y = 3x - 1$ and $y = -2x + 5$. From $y = 3x - 1$, $m_1 = 3$. From $y = -2x + 5$, $m_2 = -2$. $\tan \theta = \left|\frac{-2 - 3}{1 + (3)(-2)}\right|$ $\tan \theta = \left|\frac{-5}{1 - = 0$, then $m_1 m_2 = -1$, meaning the lines are perpendicular, and $\theta = 90^\circ$. If $m_1 = m_2$, then $m_2 - m_1 = 0$, meaning the lines are parallel, and $\theta = 0^\circ$.
Example 6: Find the acute angle between the lines $y = 3x - 1$ and $y = -2x + 5$. From $y = 3x - 1$, $m_1 = 3$. From $y = -2x + 5$, $m_2 = -2$. $\tan \theta = \left|\frac{-2 - 3}{1 + (3)(-2)}\right|$ $\tan \theta = \left|\frac{-5}{1 - 6}\right|$ $\tan \theta = \left|\frac{-5}{-5}\right|$ $\tan \theta = |1|$ $\theta = \arctan(1)$ $\theta = 45^\circ$. This section provides a detailed breakdown of the core concepts, including derivations, formulas, and worked examples essential for teaching "Coordinates geometry of straight lines." 2.
1. The Cartesian Rectangular Coordinate System The Cartesian coordinate system consists of two perpendicular number lines, the horizontal x-axis and the vertical y-axis, intersecting at a point called the origin (0,0). These axes divide the plane into four quadrants. A point in this system is represented by an ordered pair (x, y), where 'x' is the horizontal distance from the y-axis and 'y' is the vertical distance from the x-axis.
Plotting Points: To plot a point (x, y), start at the origin, move 'x' units horizontally (right for positive x, left for negative x), then move 'y' units vertically (up for positive y, down for negative y). 2.
2. Drawing and Interpreting Linear Graphs A linear graph is a straight line representing a linear relationship between two variables, typically x and y.
From an Equation: To draw a linear graph from an equation (e.g., $y = 2x + 1$):
1. Choose at least three values for x (e.g., -2, 0, 2).
2. Substitute these x-values into the equation to find the corresponding y-values.
3. Tabulate the (x, y) coordinate pairs.
4. Plot these points on the Cartesian plane.
5. Draw a straight line through the plotted points.
Interpretation: The graph visually represents how one variable changes with respect to the other. For instance, a steep line indicates a rapid change, while a flat line indicates no change. Intercepts show values when one variable is zero. 2.
3. Distance Between Two Coordinate Points Given two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$, the distance $D$ between them can be derived using the Pythagorean theorem. Consider a right-angled triangle formed by $P_1$, $P_2$, and a point $P_3(x_2, y_1)$. The horizontal leg has length $|x_2 - x_1|$, and the vertical leg has length $|y_2 - y_1|$. Applying Pythagoras' theorem ($a^2 + b^2 = c^2$): $D^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$ $D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Example 1: Calculate the distance between Lagos (L(2,3)) and Abuja (A(8,11)) on a simplified coordinate map. Let $(x_1, y_1) = (2,3)$ and $(x_2, y_2) = (8,11)$. $D = \sqrt{(8-2)^2 + (11-3)^2}$ $D = \sqrt{(6)^2 + (8)^2}$ $D = \sqrt{36 + 64}$ $D = \sqrt{100}$ $D = 10$ units. 2.
4. Midpoint of the Line Joining Two Points The midpoint $M(x_M, y_M)$ of a line segment connecting two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is found by averaging their respective x and y coordinates. $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ Example 2: Find the midpoint of the line segment joining the points A(1, 5) and B(7, -3). Let $(x_1, y_1) = (1,5)$ and $(x_2, y_2) = (7,-3)$. $x_M = \frac{1+7}{2} = \frac{8}{2} = 4$ $y_M = \frac{5+(-3)}{2} = \frac{2}{2} = 1$ The midpoint is $(4, 1)$. 2.
5. Gradient (Slope) and Intercepts of a Line Gradient (m): The gradient measures the steepness or slope of a line. It is defined as the "rise" (change in y) over the "run" (change in x) between any two points on the line. Given two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$: $m = \frac{\text{Change in y}}{\text{Change in x}} = \frac{y_2 - y_1}{x_2 - x_1}$ Interpretation: Positive gradient: Line slopes upwards from left to right.
Negative gradient: Line slopes downwards from left to right. Zero gradient ($y_2 - y_1 = 0$): Horizontal line. Undefined gradient ($x_2 - x_1 = 0$): Vertical line.
Intercepts: y-intercept (c): The point where the line crosses the y-axis. At this point, $x=0$. x-intercept: The point where the line crosses the x-axis. At this point, $y=0$.
Example 3: Find the gradient and y-intercept of the line passing through points C(2, 6) and D(5, 12). Gradient $m = \frac{12-6}{5-2} = \frac{6}{3} = 2$. To find the y-intercept, use the slope-intercept form $y = mx + c$.
Substitute one point and the gradient: Using C(2, 6): $6 = 2(2) + c \implies 6 = 4 + c \implies c = 2$. * Student Activity: Students work in groups to find the equation of a line given different conditions (e.g., point (2,3) and gradient 4; points (1,1) and (5,9)). They practice converting their answers to the general form $Ax+By+C=0$.
Step 6: Parallel and Perpendicular Lines & Angle Between Lines (10-15 mins)
Teacher Activity: Clearly state and explain the conditions for parallel lines ($m_1=m_2$) and perpendicular lines ($m_1m_2=-1$). Use visual aids or draw examples on the board. Present the formula for the angle between two lines and work through an example step-by-step. Discuss special cases ($90^\circ$ and $0^\circ$).
Student Activity: Students determine if given pairs of lines are parallel or perpendicular by calculating their gradients. They work on an example to find the angle between two intersecting lines. 3.
3. Conclusion (5 minutes)
Teacher Activity: Summarize the key formulas and concepts covered (distance, midpoint, gradient, equation of a line, parallel/perpendicular conditions, angle between lines). Reiterate the importance of coordinate geometry in various real-life Nigerian contexts (e.g., mapping, construction, business analysis). Assign homework/independent practice.
Student Activity: Students ask clarifying questions. * Note down homework assignments.
Urban Planning and Land Surveying (Nigerian Context): Coordinate geometry is extensively used by urban planners and land surveyors in Nigerian cities like Abuja, Lagos, or Port Harcourt.
Application: Plotting land boundaries, demarcating plots for housing or commercial use, planning road networks, and designing infrastructure (water pipes, electricity lines).
How it applies: Distance Formula: Used to calculate the exact lengths of property lines, road segments, or distances between proposed utility points. For instance, calculating the length of a fence needed for a rectangular plot of land given the coordinates of its corners.
Midpoint Formula: Identifying the central point for shared facilities like boreholes, transformer stations, or community centers within a defined area.
Equations of Lines: Defining exact street alignments, ensuring parallel roads maintain consistent spacing, or perpendicular intersections for efficient traffic flow. Gradients are crucial for designing drainage systems and ensuring proper water runoff. Economic Analysis and Business Forecasting: Nigerian entrepreneurs, economists, and market researchers use linear graphs to analyze trends and make predictions.
Application: Tracking profit vs. sales, demand vs. price, cost vs. production volume for local businesses (e.g., a farmer selling produce, a small-scale manufacturer, or a retail shop).
How it applies: Linear Graphs & Gradient: Plotting sales data over time to determine growth rates (gradient). For example, a local market vendor can plot weekly sales of rice bags against weeks to see if sales are increasing or decreasing and at what rate.
Equation of a Line: Deriving a linear relationship (e.g., profit = mx + c) to forecast future profits based on current sales or to determine the break-even point (where profit is zero, i.e., the x-intercept of the profit line).
Intercepts: Identifying initial costs (y-intercept for cost function) or minimum sales needed to start making a profit (x-intercept for profit function).
Navigation and Location Tracking: From GPS in smartphones to marine navigation, coordinate geometry is fundamental for determining positions and routes.
Application: Local transportation, logistics for delivery services in cities like Kano or Ibadan, and marine activities in the Niger Delta.
How it applies: Cartesian Coordinates: Representing specific locations (e.g., coordinates of a market, a specific address, or an oil rig in the ocean).
Distance Formula: Calculating the shortest path distance between two locations for delivery optimization or estimating fuel consumption.
Angle Between Lines: Determining turning angles for vehicles at intersections or the bearing of a ship's course relative to another vessel.