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How do you find the slope and intercept of $y=1.4x-7$?

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: Change of form of the given equation will give the slope, y intercept, and x-intercept of the line $y=1.4x-7$. We have it in 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 $y=1.4x-7$.
The given equation $y=1.4x-7$ is of the form $y=mx+k$. m is the slope of the line.
This gives that the slope of the line $y=1.4x-7$ is $1.4$.
Now we have to find the y intercept, and x-intercept of the same line $y=1.4x-7$.
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 converted equation is $y=1.4x-7\Rightarrow 7x-5y=35$.
Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
$\begin{align}
  & 7x-5y=35 \\
 & \Rightarrow \dfrac{7x}{35}+\dfrac{-5y}{35}=1 \\
 & \Rightarrow \dfrac{x}{5}+\dfrac{y}{-7}=1 \\
\end{align}$
Therefore, the x intercept, and y intercept of the line $y=1.4x-7$ is 5 and 7 respectively.
The intersecting points for the line $y=1.4x-7$ with the axes will be $\left( 5,0 \right)$ and $\left( 0,-7 \right)$.
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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 $.