Question

How do you write the equation $y=\dfrac{2}{3}x+1$ in standard form and identify
$a,b,c$?

Answer 1

Hint:The given equation is in the form of $y=mx+k$. m is the slope of the line. Change of form of the given equation $ax+by=c$ to find the $a,b,c$. Then, we get into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$ to find the x intercept, and y intercept of the line as p and q respectively.

Complete step by step solution:
The given equation $y=\dfrac{2}{3}x+1$ is of the form $y=mx+k$. m is the slope of the line.
This gives that the slope of the line $y=\dfrac{2}{3}x+1$ is $\dfrac{2}{3}$.
We need to convert the equation to the form of $ax+by=c$.
We multiply 3 to the both sides of the equation $y=\dfrac{2}{3}x+1$ and get
$\begin{align}
& 3y=3\left( \dfrac{2}{3}x+1 \right) \\
& \Rightarrow 3y=2x+3 \\
\end{align}$
Now we take all the variables on one side and all the constants on the other to get
$\begin{align}
& 3y=2x+3 \\
& \Rightarrow 2x-3y=-3 \\
\end{align}$
Now we get the form of $ax+by=c$. Equating the values, we get the value of $a,b,c$. Here a, b, c are the constants.
Therefore, $a=2,b=-3,c=-3$.

Note: Now we can find the y intercept, and x-intercept of the same line $2x-3y=-3$.
For this we convert the given equation 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 $2x-3y=-3$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
$\begin{align}
& 2x-3y=-3 \\
& \Rightarrow \dfrac{2x}{-3}+\dfrac{3y}{3}=1 \\
& \Rightarrow \dfrac{x}{{}^{-3}/{}_{2}}+\dfrac{y}{1}=1 \\
\end{align}$
Therefore, the x intercept, and y intercept of the line $2x-3y=-3$ is \[-\dfrac{3}{2}\] and 1
respectively.