Question

How do you write the equation of the line that passes through the point $\left( 6,-2 \right)$ and has a slope of $-\dfrac{2}{3}$?

Answer 1

Hint: We first take the general equation of a line where we have the slope and intercept form as $y=mx+c$. We put the given values of slope $m=-\dfrac{2}{3}$ and the point $\left( 6,-2 \right)$. Then we place the equation in the graph to visualise its intercept form.

Complete step-by-step answer:
We take the general equation of the line with the slope $m$ as $y=mx+c$.
It’s given that the value of the slope for our required line is $m=-\dfrac{2}{3}$.
Putting the value in the equation of $y=mx+c$, we get $y=-\dfrac{2}{3}x+c$.
As the line passes through $\left( 6,-2 \right)$.
Putting the value in the equation $y=-\dfrac{2}{3}x+c$, we get $-2=\left( -\dfrac{2}{3} \right)\times 6+c$.
We simplify the equation.
$\begin{align}
  & -2=\left( -\dfrac{2}{3} \right)\times 6+c \\
 & \Rightarrow c=4-2=2 \\
\end{align}$
This gives $c=2$.
The equation of the line becomes $y=-\dfrac{2}{3}x+2$.
To simplify the equation, we multiply both sides with 3 and get
$\begin{align}
  & 3y=3\left( -\dfrac{2}{3}x+2 \right) \\
 & \Rightarrow 3y=-2x+6 \\
 & \Rightarrow 2x+3y=6 \\
\end{align}$
The final equation of the line becomes $2x+3y=6$.

Note: For this equation $2x+3y=6$ we can convert it into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$. From the form we get that the x intercept, and y intercept of the line will be p and q respectively.
The given equation is $2x+3y=6$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
$\begin{align}
  & 2x+3y=6 \\
 & \Rightarrow \dfrac{2x}{6}+\dfrac{3y}{6}=1 \\
 & \Rightarrow \dfrac{x}{3}+\dfrac{y}{2}=1 \\
\end{align}$
The intersecting points for the line $2x+3y=6$ with the axes will be $\left( 3,0 \right)$ and $\left( 0,2 \right)$.