
How do you find the slope and intercept of $y=-7x-9$?
Answer
558k+ views
Hint: Change of form of the given equation will give the slope and y intercept of the line $y=-7x-9$. 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 y intercept of the line as q.
Complete step by step answer:
The given equation $y=-7x-9$ is of the form $y=mx+k$. Here m is the slope of the equation of the line $y=-7x-9$.
This gives that the slope of the line $y=-7x-9$ is $-7$ .
Now we have to find the y intercept, and x-intercept of the same line $y=-7x-9$.
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.
Simplifying the equation $y=-7x-9$, we get
$\begin{align}
& y=-7x-9 \\
& \Rightarrow 7x+y=-9 \\
\end{align}$
The given equation is $7x+y=-9$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
$\begin{align}
& 7x+y=-9 \\
& \Rightarrow \dfrac{7x}{-9}+\dfrac{y}{-9}=1 \\
& \Rightarrow \dfrac{x}{{}^{-9}/{}_{7}}+\dfrac{y}{-9}=1 \\
\end{align}$
Therefore, the y intercept of the line $y=-7x-9$ is -9.
The intercepting point for the line with the Y-axis is $\left( 0,-9 \right)$.
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 $.
Complete step by step answer:
The given equation $y=-7x-9$ is of the form $y=mx+k$. Here m is the slope of the equation of the line $y=-7x-9$.
This gives that the slope of the line $y=-7x-9$ is $-7$ .
Now we have to find the y intercept, and x-intercept of the same line $y=-7x-9$.
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.
Simplifying the equation $y=-7x-9$, we get
$\begin{align}
& y=-7x-9 \\
& \Rightarrow 7x+y=-9 \\
\end{align}$
The given equation is $7x+y=-9$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
$\begin{align}
& 7x+y=-9 \\
& \Rightarrow \dfrac{7x}{-9}+\dfrac{y}{-9}=1 \\
& \Rightarrow \dfrac{x}{{}^{-9}/{}_{7}}+\dfrac{y}{-9}=1 \\
\end{align}$
Therefore, the y intercept of the line $y=-7x-9$ is -9.
The intercepting point for the line with the Y-axis is $\left( 0,-9 \right)$.
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 $.
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