
How do you find the x and y intercepts for $y=-3x-9$?
Answer
559.5k+ views
Hint: Change of form of the given equation will give the x intercept and y intercept of the line $y=-3x-9$. 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. The given form is already in $y=mx+k$ to find the slope m.
Complete step-by-step solution:
The given equation $y=-3x-9$ is of the form $y=mx+k$.
Now we have to find the x intercept, and y intercept of the same line $y=-3x-9$.
For this we convert the given equation into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$. From the intercept form we get that the x intercept, and y intercept of the line will be p and q respectively.
The given equation is $y=-3x-9$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
$\begin{align}
& y=-3x-9 \\
& \Rightarrow 3x+y=-9 \\
& \Rightarrow \dfrac{3x}{-9}+\dfrac{y}{-9}=1 \\
& \Rightarrow \dfrac{x}{-3}+\dfrac{y}{-9}=1 \\
\end{align}$
Therefore, the x intercept, and y intercept of the line $y=-3x-9$ is $-3$ and $-9$ respectively.
The intercepting points for the line with the axes are $\left( -3,0 \right)$ and $\left( 0,-9 \right)$ respectively.
The form of $y=-3x-9$ is in the slope form of $y=mx+k$. This gives the slope of the line $4x+y=4$ as $-3$.
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 $y=-3x-9$ is of the form $y=mx+k$.
Now we have to find the x intercept, and y intercept of the same line $y=-3x-9$.
For this we convert the given equation into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$. From the intercept form we get that the x intercept, and y intercept of the line will be p and q respectively.
The given equation is $y=-3x-9$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
$\begin{align}
& y=-3x-9 \\
& \Rightarrow 3x+y=-9 \\
& \Rightarrow \dfrac{3x}{-9}+\dfrac{y}{-9}=1 \\
& \Rightarrow \dfrac{x}{-3}+\dfrac{y}{-9}=1 \\
\end{align}$
Therefore, the x intercept, and y intercept of the line $y=-3x-9$ is $-3$ and $-9$ respectively.
The intercepting points for the line with the axes are $\left( -3,0 \right)$ and $\left( 0,-9 \right)$ respectively.
The form of $y=-3x-9$ is in the slope form of $y=mx+k$. This gives the slope of the line $4x+y=4$ as $-3$.
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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