Equations of a Line - SS2 Mathematics Lesson Note

The general form of a linear equation in the gradient-intercept form is \(y = mx + c\), where \(m\) is the gradient of the linear graph and \(c\) is the \(y\)-intercept.

From this equation, for the \(x\)-intercept, \(y = 0\),

\[y = mx + c\]

\[0 = mx + c\]

\[- c = mx\]

\[x = \frac{- c}{m}\]

Also, from this equation, for the \(y\)-intercept, \(x = 0\),

\[y = 0 + c\]

\[y = c\]

Example 1 What is the gradient and intercepts of the linear equation \(3y = 3x + 6\)?

Solution

Expressing \(3y = 3x + 6\) in the form \(y = mx + c\),

\[\frac{3}{3}y = \frac{3}{3}x + \frac{6}{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y = x + 2\]

The gradient \(m = 1\)

The \(y\)-intercept when \(x = 0\),

\[y = c\ = 2\]

The \(x\)-intercept when \(y = 0\),

\[x = \frac{- c}{m} = \ \frac{- 2}{1} = \ - 2\]

1. Given one point \((x_{1},y_{1})\) on a linear graph, the equation of the graph can be obtained thus: \(y - y_{1} = m(x - x_{1})\)

Given two points \((x_{1},y_{1})\) and \((x_{2},y_{2})\) on a linear graph, the equation of the graph can be obtained thus: \(\frac{y - y_{1}}{x - x_{1}} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\)

Example 2 Find the equations of a straight line:

1. With gradient of \(2\), \(y\)-intercept of \(2\)

Passing through the pair of points \((3,0)\) and \((2,1)\)

With gradient of 3, passing the point \((1,3)\)

Solution

1. \(y = mx + c\)

\(y = 2x + 2\) is the required equation

1. : \(\frac{y - y_{1}}{x - x_{1}} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \ \frac{y - 0}{x - 3} = \frac{1 - 0}{2 - 3} = \ \frac{y}{x - 3} = \frac{1}{- 1}\)

\[- y = x - 3\]

\(y = - x + 3\) is the required equation

1. \(y - y_{1} = m(x - x_{1})\)

\[y - 3 = 3(x - 1)\]

\[y - 3 = 3x - 3\]

\[y = 3x - 3 + 3\]

\(y = 3x\) is the required equation