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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 $\left( 0,-7 \right)$.**

**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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