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IB Mathematics AI SL Different forms of the equation of a straight line MAI Study Notes - New Syllabus

IB Mathematics AI SL Operations with numbers Study Notes

LEARNING OBJECTIVE

  • Different forms of the equation of a straight line.

Key Concepts: 

  • Equations of a Straight Line

MAI HL and SL Notes – All topics

 LINES (or LINEAR FUNCTIONS)

♦ BASIC NOTIONS ON COORDINATE GEOMETRY

Given two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \):

♦ The gradient or slope of line segment AB is given by:

\(m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{\Delta y}{\Delta x}\)

This indicates the inclination of the line segment AB. As we move along the positive direction of the x-axis:

If the line segment is increasing (⁄), then \( m > 0 \).
If the line segment is decreasing (\ ), then \( m < 0 \).
If the line segment is horizontal (—), then \( m = 0 \).
If the line segment is vertical (|), then \( m \) is not defined.

♦ The distance between A and B is given by:

\(d_{AB} = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}\)

♦ The coordinates of the midpoint \( M(x, y) \) of the line segment AB are given by:

\(x = \frac{x_1 + x_2}{2}, \quad y = \frac{y_1 + y_2}{2}\)

Example

a) Given two points \( A(1, 4) \) and \( B(7, 12) \):

b) Given two points \( A(1, 8) \) and \( B(5, 8) \):

c) Given two points \( A(1, 5) \) and \( B(1, 7) \):

▶️Answer/Explanation

Solution:

(a)  The slope of the line segment AB is:

\(m = \frac{12 – 4}{7 – 1} = \frac{8}{6} = \frac{4}{3}\)

The distance between them is:

\(d = \sqrt{(7 – 1)^2 + (12 – 4)^2} = \sqrt{36 + 64} = 10\)

 The midpoint is:

\(M\left(\frac{1 + 7}{2}, \frac{4 + 12}{2}\right) = M(4, 8)\)

(b) The slope is \( m = 0 \) (horizontal).
 The distance is \( d = 5 – 1 = 4 \).
 The midpoint is \( M(3, 8) \).

(c) The slope \( m \) is not defined (vertical).
The distance is \( d = 7 – 5 = 2 \).
The midpoint is \( M(1, 6) \).

THE EQUATION OF A LINE

♦ Equation of a (straight) line:

\(y = mx + c\)

\( m = \) gradient or slope.
\( c = y \)-intercept.

♦ NOTICE:
A horizontal line has equation \( y = c \) (slope \( m = 0 \)).
A vertical line has equation \( x = c \) (no slope).
(A vertical line is not a function, which is why the equation \( x = c \) is not a particular case of \( y = mx + c \).)

Look at the graphs of two lines:
\( L_1: y = 2x \).
\( L_2: y = -2x \).

The slope shows the rise of the line per each unit:
 Line \( L_1 \): Slope is 2 (\( y \) increases 2 units per each \( x \)-unit).
 Line \( L_2 \): Slope is \(-2\) (\( y \) decreases 2 units per each \( x \)-unit).

In both cases, \( c = 0 \) (the function passes through the origin).

Example

Look at the graphs of two lines: \( L_1: y = 2x + 3 \) and \( L_2: y = -2x + 3 \)

▶️Answer/Explanation

Solution:


Line \( L_1 \): slope is 2
Line \( L_2 \): slope is -2
In both cases the \( y \)-intercept is 3

Example

Look at the graphs of two lines: \( L_1: y = 5 \) and \( L_2: x = 5 \)

▶️Answer/Explanation

Solution:

♦PARALLEL AND PERPENDICULAR LINES

Consider two lines: \( L_1: y = m_1 x + c_1 \) and \( L_2: y = m_2 x + c_2 \)

Parallel lines:

\( L_1 // L_2 \) if \( m_1 = m_2 \)

Perpendicular lines:

\( L_1 \perp L_2 \) if \( m_2 = -1 / m_1 \)

For example,

The lines \( y = 3x + 5 \) and \( y = 3x + 8 \) are parallel

The lines \( y = 3x + 5 \) and \( y = -\frac{1}{3}x + 8 \) are perpendicular

♦AN ALTERNATIVE FORMULA FOR A LINE

A more general formula for a line is

Equation of a line:

\( Ax + By = C \)

If \( B \neq 0 \), we can solve for \( y \) and obtain the form \( y = mx + c \)
If \( B = 0 \), we obtain a vertical line of the form \( x = c \)
If \( A = 0 \), we obtain a horizontal line of the form \( y = c \)

From \( Ax + By = C \) into the usual form
The line \( 2x + 3y = 5 \) may be expressed as \( 3y = -2x + 5 \) and finally

\( y = \frac{-2}{3}x + \frac{5}{3} \)

From the usual form into \( Ax + By = C \)
a) The line \( y = -3x + 7 \) may be expressed as

\( 3x + y = 7 \)

b) The line \( y = \frac{1}{2}x + \frac{2}{3} \) may be expressed as

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

We usually require the coefficients \( A, B, C \) to be integers.

Multiplying by 6 we obtain

\( -3x + 6y = 4 \)

c) The line \( y = 5 \) may be expressed as \( 0x + y = 5 \)
d) The line \( x = 5 \) may be expressed as \( x + 0y = 5 \)

♦A POINT AND A SLOPE

The line which passes through point \( P(x_0, y_0) \)  has slope \( m \) is given by

\( y – y_0 = m(x – x_0) \)

Example

The line which passes through point \( P(1, 2) \), with slope \( m = 3 \) is

\( y – 2 = 3(x – 1) \)

 Express in the form \( y = mx + c \)

▶️Answer/Explanation

Solution:

\( y – 2 = 3(x – 1) \Leftrightarrow y = 3x – 3 + 2 \Leftrightarrow y = 3x – 1 \)

Express in the form \( ax + by = c \) or \( ax + by + c = 0 \)

\( y = 3x – 1 \Leftrightarrow 3x – y = 1 \quad \text{or} \quad 3x – y – 1 = 0 \)

♦ TWO POINTS

The line which passes through the points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) has slope

\( m = \frac{\Delta y}{\Delta x} = \frac{y_2 – y_1}{x_2 – x_1} \)

and its equation is again given by the formula

\( y – y_1 = m(x – x_1) \)

Example

Find the line which passes through the points \( P(1, 2) \) and \( Q(4, 7) \).
Express your answer in the form \( ax + by = c \) where \( a, b, c \in Z \) (integers).

▶️Answer/Explanation

Solution:

The slope is

\( m = \frac{\Delta y}{\Delta x} = \frac{7 – 2}{4 – 1} = \frac{5}{3} \)

The equation of the line is

\( y – 2 = \frac{5}{3}(x – 1)\)
\( \Leftrightarrow 3y – 6 = 5(x – 1)\)
\( \Leftrightarrow 3y – 6 = 5x – 5\)
\( -5x + 3y = 1 \)

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