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Chapter 7: Straight Line Graphs

7.1 Equation of a Straight Line: \(y = mx + c\)

General form:
The equation of a straight line can be written as: \[ y = mx + c \] where:

Proof: The gradient formula

The gradient \(m\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Derivation:
The gradient measures the rate of change of \(y\) with respect to \(x\). For a straight line, this rate is constant.
Given two points on the line, the change in \(y\) is \(\Delta y = y_2 - y_1\), and the change in \(x\) is \(\Delta x = x_2 - x_1\).
The ratio \(\frac{\Delta y}{\Delta x}\) is the gradient.

Alternative form (point-gradient form):
The equation of a line passing through \((x_1, y_1)\) with gradient \(m\) is: \[ y - y_1 = m(x - x_1) \] This is derived directly from the definition of gradient.

General form:
\(ax + by + c = 0\) can be rearranged to \(y = mx + c\) by solving for \(y\).

Worked Example 7.1 (Finding gradient and intercept – non-calculator)
Find the gradient and y-intercept of the line \(3x - 2y + 6 = 0\).
Solution:
Rearrange to \(y = mx + c\): \[ -2y = -3x - 6 \] \[ y = \frac{3}{2}x + 3 \] Answer: Gradient \(m = \frac{3}{2}\), y-intercept \(c = 3\).
Worked Example 7.2 (Equation from two points – non-calculator)
Find the equation of the line passing through \((2, 3)\) and \((5, 9)\).
Solution:
Gradient: \[ m = \frac{9 - 3}{5 - 2} = \frac{6}{3} = 2 \] Using point-gradient form with \((2, 3)\): \[ y - 3 = 2(x - 2) \] \[ y - 3 = 2x - 4 \] \[ y = 2x - 1 \] Answer: \(y = 2x - 1\)
Worked Example 7.3 (Mauritian context – cost of taxi)
A taxi in Port Louis charges a fixed fee plus a rate per kilometre. A journey of 5 km costs Rs 250, and a journey of 10 km costs Rs 400. Find the fixed fee and the rate per km.
Solution:
Let fixed fee = \(c\), rate per km = \(m\).
Then cost \(y = mx + c\), where \(x\) is distance.

Points: \((5, 250)\) and \((10, 400)\).

Gradient: \[ m = \frac{400 - 250}{10 - 5} = \frac{150}{5} = 30 \] Using \((5, 250)\): \(250 = 30(5) + c \Rightarrow 250 = 150 + c \Rightarrow c = 100\)

Answer: Fixed fee Rs 100, rate Rs 30 per km.

7.2 Gradient and Intercept from an Equation

Given a linear equation, we can find the gradient and y-intercept by rewriting it in the form \(y = mx + c\).

For a line in the form \(ax + by + c = 0\): \[ by = -ax - c \Rightarrow y = -\frac{a}{b}x - \frac{c}{b} \] So:

Worked Example 7.4 (Identifying gradient and intercept)
Find the gradient and intercept of \(2x + 3y - 12 = 0\).
Solution:
\[ 3y = -2x + 12 \] \[ y = -\frac{2}{3}x + 4 \] Answer: Gradient \(= -\frac{2}{3}\), y-intercept \(= 4\).

7.3 Mid-point and Length of a Line Segment

Mid-point formula:
The midpoint of the line segment joining \((x_1, y_1)\) and \((x_2, y_2)\) is: \[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]

Proof: Mid-point formula

The midpoint is the average of the x-coordinates and the average of the y-coordinates.
This follows from the fact that the midpoint divides the segment in the ratio \(1:1\), and we are finding the point that is halfway along the segment.

Distance (length) formula:
The length of the line segment joining \((x_1, y_1)\) and \((x_2, y_2)\) is: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] This is derived from the Pythagorean theorem.

Proof: Distance formula

Consider the right-angled triangle formed by the two points.
The horizontal distance is \(\Delta x = x_2 - x_1\).
The vertical distance is \(\Delta y = y_2 - y_1\).
By Pythagoras: \(d^2 = (\Delta x)^2 + (\Delta y)^2\).
Taking square roots gives \(d = \sqrt{(\Delta x)^2 + (\Delta y)^2}\).
Worked Example 7.5 (Mid-point and length – non-calculator)
Find the midpoint and length of the segment joining \(A(2, 3)\) and \(B(8, 11)\).
Solution:
Midpoint: \[ \left(\frac{2 + 8}{2}, \frac{3 + 11}{2}\right) = (5, 7) \] Length: \[ \sqrt{(8 - 2)^2 + (11 - 3)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \] Answer: Midpoint \((5, 7)\), length \(10\).
Worked Example 7.6 (Mauritian context – distance between towns)
On a map of Mauritius, Port Louis is at \((2, 5)\) and Mahébourg is at \((14, 11)\), where units are in km. Find the distance between the two towns.
Solution:
\[ d = \sqrt{(14 - 2)^2 + (11 - 5)^2} = \sqrt{12^2 + 6^2} = \sqrt{144 + 36} = \sqrt{180} = 6\sqrt{5} \approx 13.42 \text{ km} \] Answer: Approximately 13.42 km.

7.4 Parallel and Perpendicular Lines

Parallel lines:
Two lines are parallel if they have the same gradient: \[ m_1 = m_2 \]

Perpendicular lines:
Two lines are perpendicular if the product of their gradients is \(-1\): \[ m_1 \times m_2 = -1 \]

Proof: Perpendicular lines condition

Consider two lines with gradients \(m_1\) and \(m_2\). The angle \(\theta\) of a line with the x-axis satisfies \(m = \tan \theta\).

For perpendicular lines, \(\theta_2 = \theta_1 + 90^\circ\).
So \(m_2 = \tan(\theta_1 + 90^\circ) = -\frac{1}{\tan \theta_1} = -\frac{1}{m_1}\).

Hence \(m_1 m_2 = -1\).
Worked Example 7.7 (Parallel lines – non-calculator)
Find the equation of the line parallel to \(y = 3x - 2\) that passes through \((1, 5)\).
Solution:
Parallel means same gradient: \(m = 3\).

Using point-gradient form: \[ y - 5 = 3(x - 1) \] \[ y - 5 = 3x - 3 \] \[ y = 3x + 2 \] Answer: \(y = 3x + 2\)
Worked Example 7.8 (Perpendicular lines – non-calculator)
Find the equation of the line perpendicular to \(y = -\frac{1}{2}x + 4\) that passes through \((3, -1)\).
Solution:
Gradient of given line: \(m_1 = -\frac{1}{2}\).

Perpendicular gradient: \(m_2 = -\frac{1}{m_1} = -\frac{1}{-1/2} = 2\).

Using point-gradient form: \[ y - (-1) = 2(x - 3) \] \[ y + 1 = 2x - 6 \] \[ y = 2x - 7 \] Answer: \(y = 2x - 7\)

7.5 Perpendicular Bisectors

The perpendicular bisector of a line segment is:

Steps to find the perpendicular bisector:

  1. Find the midpoint of the segment
  2. Find the gradient of the segment
  3. Find the negative reciprocal (perpendicular gradient)
  4. Use point-gradient form to write the equation
Worked Example 7.9 (Perpendicular bisector – non-calculator)
Find the equation of the perpendicular bisector of the segment joining \(A(2, 5)\) and \(B(8, 9)\).
Solution:
Midpoint: \[ \left(\frac{2 + 8}{2}, \frac{5 + 9}{2}\right) = (5, 7) \] Gradient of AB: \[ m_{AB} = \frac{9 - 5}{8 - 2} = \frac{4}{6} = \frac{2}{3} \] Perpendicular gradient: \[ m = -\frac{3}{2} \] Equation through \((5, 7)\) with gradient \(-\frac{3}{2}\): \[ y - 7 = -\frac{3}{2}(x - 5) \] \[ y - 7 = -\frac{3}{2}x + \frac{15}{2} \] \[ y = -\frac{3}{2}x + \frac{15}{2} + 7 \] \[ y = -\frac{3}{2}x + \frac{15}{2} + \frac{14}{2} = -\frac{3}{2}x + \frac{29}{2} \] Answer: \(y = -\frac{3}{2}x + \frac{29}{2}\)

7.6 Linearisation of Non-Linear Relationships

Purpose: To convert non-linear relationships into straight line form so that unknown constants can be found from the gradient and intercept.

Case 1: \(y = ax^n\)

Take logarithms (natural or base 10): \[ \ln y = \ln(a x^n) = \ln a + n \ln x \] Let:

So plotting \(\ln y\) against \(\ln x\) gives a straight line with gradient \(n\) and intercept \(\ln a\).

Case 2: \(y = Ab^x\)

Take logarithms: \[ \ln y = \ln(A b^x) = \ln A + x \ln b \] Let:

So plotting \(\ln y\) against \(x\) gives a straight line with gradient \(\ln b\) and intercept \(\ln A\).

Worked Example 7.10 (Linearising \(y = ax^n\))
Experimental data suggests that \(y\) and \(x\) are related by \(y = a x^n\). A graph of \(\ln y\) against \(\ln x\) gives a straight line with gradient 2.5 and intercept 3. Find \(a\) and \(n\).
Solution:
Gradient = \(n = 2.5\)
Intercept = \(\ln a = 3 \Rightarrow a = e^3 \approx 20.09\)

Answer: \(y = 20.09 x^{2.5}\)
Worked Example 7.11 (Mauritian context – population growth)
The population \(P\) of a town in Mauritius is recorded over time \(t\) (in years). The data follows \(P = A b^t\). A graph of \(\ln P\) against \(t\) gives a straight line with gradient 0.04 and intercept 6. Find the population model.
Solution:
Intercept = \(\ln A = 6 \Rightarrow A = e^6 \approx 403.43\)
Gradient = \(\ln b = 0.04 \Rightarrow b = e^{0.04} \approx 1.0408\)

Answer: \(P = 403.43(1.0408)^t\)

Chapter 7 Summary

Concept Key points
Line equation \(y = mx + c\)
Gradient \(m = \frac{y_2 - y_1}{x_2 - x_1}\)
Midpoint \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
Length \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
Parallel \(m_1 = m_2\)
Perpendicular \(m_1 \times m_2 = -1\)
Perpendicular bisector Through midpoint, gradient negative reciprocal
Linearisation (\(y = ax^n\)) Plot \(\ln y\) vs \(\ln x\)
Linearisation (\(y = Ab^x\)) Plot \(\ln y\) vs \(x\)

Exercises – Chapter 7

Easy (Drill – Non-Calculator)

  1. Find the gradient and y-intercept of:
    (a) \(y = 4x - 3\)
    (b) \(2x + 5y - 10 = 0\)
    (c) \(3x - 4y + 8 = 0\)
  2. Find the equation of the line passing through:
    (a) \((1, 2)\) and \((4, 8)\)
    (b) \((-2, 3)\) and \((4, -1)\)
  3. Find the midpoint and length of the segment joining:
    (a) \((0, 0)\) and \((6, 8)\)
    (b) \((2, -1)\) and \((8, 7)\)

Medium (Examination Style)

  1. Find the equation of the line through \((3, -2)\) that is:
    (a) parallel to \(y = 2x - 5\)
    (b) perpendicular to \(y = -\frac{1}{3}x + 4\)
  2. Find the perpendicular bisector of the segment joining \(A(1, 3)\) and \(B(7, 9)\).
  3. Mauritian context: A taxi company charges a fixed fee plus a rate per km. A 6 km journey costs Rs 280, and a 12 km journey costs Rs 460. Find the fixed fee and rate per km.
  4. Experimental data follows \(y = ax^n\). A graph of \(\ln y\) against \(\ln x\) has gradient 1.5 and intercept 2. Find \(a\) and \(n\).

Hard (Challenge for A*)

  1. The line \(y = 2x - 3\) is parallel to the line \(y = mx + c\), which passes through \((4, 5)\). Find \(m\) and \(c\).
  2. The line \(y = 3x + 1\) is perpendicular to the line \(y = mx + c\), which passes through \((2, -1)\). Find \(m\) and \(c\).
  3. Mauritian context: The value of a car in Mauritius depreciates according to \(V = Ab^t\), where \(t\) is the age in years.
    (a) A graph of \(\ln V\) against \(t\) has gradient \(-0.15\) and intercept 9. Find the model.
    (b) What was the initial value of the car?
    (c) What will the car be worth after 5 years?
    (d) Sketch the graph of \(\ln V\) against \(t\).
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