Question

How do you write $y+3=-\dfrac{1}{3}\left( 2x+6 \right)$ in slope-intercept form?

Answer 1

Hint: Change of form of the given equation will give the slope, y intercept, and x-intercept of the line $y+3=-\dfrac{1}{3}\left( 2x+6 \right)$. 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 $y+3=-\dfrac{1}{3}\left( 2x+6 \right)$.
The simplified form will be
$\begin{align}
  & y+3=-\dfrac{1}{3}\left( 2x+6 \right) \\
 & \Rightarrow 3y+9=-2x-6 \\
 & \Rightarrow 2x+3y=-15 \\
\end{align}$
The equation $2x+3y=-15$ 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
$\begin{align}
  & 2x+3y=-15 \\
 & \Rightarrow y=\dfrac{-2}{3}x-5 \\
\end{align}$
This gives that the slope of the line $y+3=-\dfrac{1}{3}\left( 2x+6 \right)$ is $\dfrac{-2}{3}$.
Now we have to find the y intercept, and x-intercept of the same line $2x+3y=-15$.
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 $2x+3y=-15$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
$\begin{align}
  & 2x+3y=-15 \\
 & \Rightarrow \dfrac{2x}{-15}+\dfrac{3y}{-15}=1 \\
 & \Rightarrow \dfrac{x}{{}^{-15}/{}_{2}}+\dfrac{y}{-5}=1 \\
\end{align}$
Therefore, the x intercept, and y intercept of the line $2x+3y=-15$ is $-\dfrac{15}{2}$ and 5 respectively.
The intersecting points for the line $2x+3y=-15$ with the axes will be $\left( -\dfrac{15}{2},0 \right)$ and $\left( 0,-5 \right)$.

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 $.