Straight Lines

9.1  Straight Lines
  • For a linear function, \(y=mx+c\)\(m\) is the gradient and \(c\) is the \(y\)-intercept of the straight line.
  • The graph of a linear function, \(y=mx+c\) is a straight line.

Equation of a straight line:
where \(m\) is the gradient and \(c\) is the \(y\)-intercept.
  • The graph of function \(y=h\) is a straight line parallel to \(x\)-axis.
  • The graph of function \(x=h\) is a straight line parallel to \(y\)-axis.

Determine the gradient and \(y\)-intercept of the straight line \(y=4x+9\).


Noted that the general equation of a straight line is \(y=mx+c\).

We need to compare \(y=4x+9\) with \(y=mx+c\).

So, \(m=4\) and \(c=9\).

Thus, the gradient is \(4\) and the \(y\)-intercept is \(9\).

Relationship between the equation of straight lines in the form of \(ax+by=c\)\(\dfrac{x}{a} +\dfrac{y}{b}=1\) and \(y=mx+c\):
  • The values of \(x\)-intercept, \(y\)-intercept and the gradient of these three straight lines are the same.
  • Produce the same straight line graph if the values of \(x\)-intercept and \(y\)-intercept are the same.
  • The straight line equation, \(y=mx+c\), can also be written in the form of \(ax+by=c\) and \(\dfrac{x}{a} +\dfrac{y}{b}=1\) where \(a\neq0\) and \(b\neq0\).

Change the equation \(4x+3y=12\) to the form of \(\dfrac{x}{a} +\dfrac{y}{b}=1\) and \(y=mx+c\).


i) For \(\dfrac{x}{a} +\dfrac{y}{b}=1\)

\(\begin{aligned} 4x+3y&=12 \\\\\dfrac{4x}{12}+\dfrac{3y}{12}&=\dfrac{12}{12} \\\\\dfrac{x}{3}+\dfrac{y}{4}&=1. \end{aligned}\)


ii) For \(y=mx+c\)

\(\begin{aligned} 4x+3y&=12 \\\\3y&=-4x+12 \\\\ \dfrac{3y}{3}&=\dfrac{-4x}{3}+\dfrac{12}{3} \\\\y&=-\dfrac{4}{3}x +4. \end{aligned}\)

The points on a straight line and the equation of the line:
  • Points on a straight line or points that the straight line passes through will satisfy the equation of a straight line.
  • Points that do not lie on a straight line will not satisfy the equation. 

Determine whether point \(P\) satisfy the given equation.

\(y= 3x+2\)\(P (2,8)\)


On the left side of the equation,


Meanwhile, on the right side of the equation is,

\(\begin{aligned} 3x+2&= 3(2)+2 \\\\&=8. \end{aligned}\)

We can see that the value on the left and right sides of the equation is equal.

\(P (2,8)\) lies on the straight line \(y= 3x+2\).

Thus, point \(P\) satisfy the equation \(y= 3x+2\).

  • For points that lie on \(x\)-axis its value of \(y\)-coordinate is \(0\).
  • For points that lie on \(y\)-axis its value of \(x\)-coordinate is \(0\).

Gradient of a straight line, \(m\)


The gradients of parallel lines:
  • Straight lines that have the same gradient are parallel.

Determine whether \(y=3x+8\) is parallel to \(6y=3x-9\).


Noted that the equation of a straight line is \(y=mx+c\), where \(m\) is the gradient and \(c\) is the \(y\)-intercept.

For \(y=3x+8\), the gradient is \(m=3\).

For \(6y=3x-9\),

\(\begin{aligned} 6y&=3x-9 \\\\y&=\dfrac{3x}{6}-\dfrac{9}{6} \\\\y&=\dfrac{1}{2}x-\dfrac{3}{2}. \end{aligned}\)

So, the gradient is \(m=\dfrac{1}{2}\).

The gradient of both straight lines is not equal.

Thus, \(y=3x+8\) and \(6y=3x-9\) is not parallel.

The equation of a straight line:
  1. Determine the value of gradient, \(m\).
  2. Determine a point which the straight line passes through or a point that lies on the straight line.
  3. Substitute the gradient, \(m\), the \(x\)-coordinate and the \(y\)-coordinate from the point into the equation \(y=mx+c\) to determine the value of \(c\), that is the \(y\)-intercept.
  4. Substitute the gradient value and \(y\)-intercept value specified in the equation of the straight line \(y=mx+c\).
The point of intersection of two straight lines:

Can be determined by the following methods;

  • Graphical method
  • Substitution method
  • Elimination method

Determine the point of intersection of the straight lines \(2x+y=5\) and \(x+2y=1\).


By using the substitution method;

\(2x+y=5\)  ------\(\boxed{1}\)

\(x+2y=1\)  ------\(\boxed{2}\)

From \(\boxed{1}\)

\(y=5-2x\)  ------\(\boxed{3}\)

Substitute \(\boxed{3}\) in \(\boxed{2}\),

\(\begin{aligned} x+2(5-2x)&=1 \\\\x+10-4x&=1 \\\\x-4x&=1-10 \\\\-3x&=-9 \\\\x&=3. \end{aligned}\)

Substitute \(x=3\) in \(\boxed{3}\),

\(\begin{aligned} y&=5-2(3) \\\\y&=-1. \end{aligned}\)

Hence, the point of intersection is \((3,-1)\).