How do you write the equation in a slope intercept form when we are given the points as $\left( 1,4 \right)$ and $\left( 2,-5 \right)$?

Hint: We have to first find the slope of the given equation with points $\left( 1,4 \right)$ and $\left( 2,-5 \right)$. We use that to find the equation of line as $\dfrac{y-b}{b-d}=\dfrac{x-a}{a-c}$ for points $\left( a,b \right)$ and $\left( c,d \right)$. We put the values for $\left( 1,4 \right)$ and $\left( 2,-5 \right)$ to find the line.

Complete step-by-step solution:
We have been given two points $\left( 1,4 \right)$ and $\left( 2,-5 \right)$.
We need to find the slope intercept form and also the equation of the line.
We know that the slope for the line with given points $\left( a,b \right)$ and $\left( c,d \right)$ will be $\dfrac{d-b}{c-a}$.
Now we assume the value of the slope as $m=\dfrac{d-b}{c-a}$.
We can form the equation from this value of slope where we find the equation using any one of the given points.
So, the equation becomes $y=mx+c$.
So, we put the value of m and $\left( a,b \right)$ to get $c=b-a\times \dfrac{d-b}{c-a}$.
The simplified form for points $\left( a,b \right)$ and $\left( c,d \right)$ becomes $\dfrac{y-b}{b-d}=\dfrac{x-a}{a-c}$.
For our given problem we use the points $\left( 1,4 \right)$ and $\left( 2,-5 \right)$ to get the equation as
$\begin{align}
  & \dfrac{y-4}{4-\left( -5 \right)}=\dfrac{x-1}{1-2} \\
 & \Rightarrow \dfrac{y-4}{9}=\dfrac{x-1}{-1} \\
 & \Rightarrow y-4=9-9x \\
 & \Rightarrow 9x+y=13 \\
& \Rightarrow y= -9x+13 \\
\end{align}$
The line is $y= -9x+13 $.

Note: we need to remember that the intercept of the line can be found by converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
$\begin{align}
  & 9x+y=13 \\
 & \Rightarrow \dfrac{9x}{13}+\dfrac{y}{13}=1 \\
 & \Rightarrow \dfrac{x}{{}^{13}/{}_{9}}+\dfrac{y}{13}=1 \\
\end{align}$
Therefore, the x intercept, and y intercept of the line $9x+y=13$ is $\dfrac{13}{9}$ and 13 respectively.
The intersecting points for the line $9x+y=13$ with the axes will be $\left( \dfrac{13}{9},0 \right)$ and $\left( 0,13 \right)$.