Question

How do you find the slope of $5x+2y=-20$?

Answer 1

Hint: Change of form of the given equation will give the slope, y intercept, and x-intercept of the line $5x+2y=-20$. 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.

Complete step by step answer:
The given equation $5x+2y=-20$ 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
$\begin{align}
  & 5x+2y=-20 \\
 & \Rightarrow 2y=-5x-20 \\
 & \Rightarrow y=\dfrac{-5x-20}{2}=\dfrac{-5x}{2}-10 \\
\end{align}$
This gives that the slope of the line $5x+2y=-20$ is $-\dfrac{5}{2}$.
Now we have to find the y intercept, and x-intercept of the same line $5x+2y=-20$.
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 $5x+2y=-20$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
$\begin{align}
  & 5x+2y=-20 \\
 & \Rightarrow \dfrac{5x}{-20}+\dfrac{2y}{-20}=1 \\
 & \Rightarrow \dfrac{x}{-4}+\dfrac{y}{-10}=1 \\
\end{align}$
Therefore, the x intercept, and y intercept of the line $5x+2y=-20$ is $4$ and $10$ respectively. The intercepting points for the line with the axes are $\left( -4,0 \right)$ and $\left( 0,-10 \right)$ respectively.

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 $.