
How do you find the slope, y intercept, and x-intercept of the line $2x+y=5$?
Answer
573.9k+ views
Hint: Change of form of the given equation will give the slope, y intercept, and x-intercept of the line $2x+y=5$. 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 solution:
The given equation $2x+y=5$ 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
$\begin{align}
& 2x+y=5 \\
& \Rightarrow y=-2x+5 \\
\end{align}$
This gives that the slope of the line $2x+y=5$ is -2.
Now we have to find the y intercept, and x-intercept of the same line $2x+y=5$.
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 $2x+y=5$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
$\begin{align}
& 2x+y=5 \\
& \Rightarrow \dfrac{2x}{5}+\dfrac{y}{5}=1 \\
& \Rightarrow \dfrac{x}{{}^{5}/{}_{2}}+\dfrac{y}{5}=1 \\
\end{align}$
Therefore, the x intercept, and y intercept of the line $2x+y=5$ is $\dfrac{5}{2}$ and 5 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 $.
Complete step-by-step solution:
The given equation $2x+y=5$ 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
$\begin{align}
& 2x+y=5 \\
& \Rightarrow y=-2x+5 \\
\end{align}$
This gives that the slope of the line $2x+y=5$ is -2.
Now we have to find the y intercept, and x-intercept of the same line $2x+y=5$.
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 $2x+y=5$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
$\begin{align}
& 2x+y=5 \\
& \Rightarrow \dfrac{2x}{5}+\dfrac{y}{5}=1 \\
& \Rightarrow \dfrac{x}{{}^{5}/{}_{2}}+\dfrac{y}{5}=1 \\
\end{align}$
Therefore, the x intercept, and y intercept of the line $2x+y=5$ is $\dfrac{5}{2}$ and 5 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 $.
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