How do you find the slope, x intercept and y intercept of $3x+4y=6$?

Hint: Change of form of the given equation will give the slope, y intercept, and x-intercept of the line $3x+4y=6$. We change it to the form of $y=mx+k$ to find the slope m. 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. then we place the line on the graph based on that

Complete step by step solution:
We are taking the general equation of line to understand the slope and the intercept form of the line $3x+4y=6$.
The given equation $3x+4y=6$ is of the form $ax+by=c$. Here a, b, c are the constants.
We convert the form to $y=mx+k$. m is the slope of the line.
So, converting the equation we get
$\begin{align}
  & 3x+4y=6 \\
 & \Rightarrow y=\dfrac{6-3x}{4}=-\dfrac{3}{4}x+\dfrac{3}{2} \\
\end{align}$
This gives that the slope of the line $3x+4y=6$ is $-\dfrac{3}{4}$.
Now we have to find the y intercept, and x-intercept of the same line $3x+4y=6$.
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 $3x+4y=6$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
$\begin{align}
  & 3x+4y=6 \\
 & \Rightarrow \dfrac{3x}{6}+\dfrac{4y}{6}=1 \\
 & \Rightarrow \dfrac{x}{2}+\dfrac{y}{{}^{3}/{}_{2}}=1 \\
\end{align}$
Therefore, the x intercept, and y intercept of the line $3x+4y=6$ is 2 and $\dfrac{3}{2}$ respectively.
The intersecting points for the line $3x+4y=6$ with the axes will be $\left( 2,0 \right)$ and $\left( 0,\dfrac{3}{2} \right)$.


Note: A line parallel to the X-axis does not intersect the X-axis at any finite distance and hence we cannot get any finite x-intercept of such a line. Same goes for lines parallel to the Y-axis. In case of slope of a line the range of the slope is 0 to $\infty $.