Question

How do you draw the graph of equation $y=\dfrac{6}{5}x+5$ using the slope-intercept form? \[\]

Answer 1

Hint: We recall the definition of slope and intercept of a line. We recall that the slope-intercept form of line is given by $y=mx+c$ where $m$ is the slope and $c$ is the intercept. We first plot $C\left( 0,c \right)$ and distance $\left| c \right|$ from the origin $O\left( 0,0 \right)$. We find the other point $A\left( a,0 \right)$ by putting $y=0$ in the given line and then finding the value of $x$. \[\]

Complete step-by-step solution:
 We know that the intercept of a line is the $y-$coordinate of the point where it crosses the $y-$axis. We also know that the slope of the line is the degree of tilde of the line with a positive $x-$axis. It is given by the tangent of the angle the line makes with the positive $x-$axis. The general slope-intercept form of the line with slope $m$ and intercept $c$ is given by
\[y=mx+c\]
We are given in the question the equation of the line
\[y=\dfrac{6}{5}x+5\]
We compare the equation of above line with general slope-point form to have slope $m=\dfrac{6}{5}$ and intercept $c=5$. We need two points to draw a line. So we first plot the point $C\left( 0,5 \right)$ at distance 5 from the origin $O\left( 0,0 \right)$ where the given line cuts the $y-$axis. So we have
\[OC=5,m=\dfrac{6}{5}\]
Now let us put $y=0$ in the given equation to have;
\[\begin{align}
  & 0=\dfrac{6}{5}x+5 \\
 & \Rightarrow \dfrac{6}{5}x=-5 \\
 & \Rightarrow x=\dfrac{-25}{6} \\
\end{align}\]
So the other point of the equation is $A\left( \dfrac{-25}{6},0 \right)$. We join the points $A\left( \dfrac{-25}{6},0 \right),C\left( 0, 5 \right)$ to get the required line. \[\]

Note: We note that the tangent of the angle is the ratio of the opposite side to the adjacent side (excluding hypotenuse) in a right-angle triangle. So in right-angled triangle COA we have$\tan \theta =\tan \left( \angle CAO \right)=\dfrac{OC}{OA}=\dfrac{5}{\left| \dfrac{-25}{6} \right|}=\dfrac{6}{5}$. Hence the drawn line has the required slope. We also note that if the slope is negative the line will subtend the obtuse angle and if the slope is positive the line will subtend acute angle with positive $x-$axis.