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: $$y=mx+c$$ 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.
 Example 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 $$b\neq0$$.
 Example 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.
 Example Determine whether point $$P$$ satisfy the given equation. $$y= 3x+2$$, $$P (2,8)$$ On the left side of the equation, $$y=8$$. 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, $$m$$ $$m=-\dfrac{y-\text{intercept}}{x-\text{intercept}}$$
 The gradients of parallel lines: Straight lines that have the same gradient are parallel. Example 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: Determine the value of gradient, $$m$$. Determine a point which the straight line passes through or a point that lies on the straight line. 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. 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 Example 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 $$(3,-1)$$.