How do you write the equation in standard form given $m = 4$ and $(4,3)$ ?

Hint: First, we will identify the points and slope in the given problem.
$({x_1},{y_1})$ is the passing point in the equation. Find the point ${x_1}$ and ${y_1}$ .
And then, substitute the points and slope in the point-slope form of the equation. The equation is as follows:
$y - {y_1} = m(x - {x_1})$
After that, we will group variables and constants on both sides.
And finally, we will find the equation in the form of $\operatorname{Ax} + By = C$ , where $\operatorname{A} $ should be positive. If it is not, we will multiply the entire equation by $ - 1$ .

Complete step-by-step solution:
The given information is –
$m = 4$
And the point is $(4,3)$
We are given a point and a slope, so we can use point-slope form;
The point-slope form of equation is,
$y - {y_1} = m(x - {x_1})$
The point is,
${x_1}$ - the point is $4$
${y_1}$ - the point is $3$
We plug in the numbers we know and simplify until we get into standard form.
Now, we will apply the values of $m,{x_1},{y_1}$ in the equation, hence we get,
$\Rightarrow$$y - 3 = 4(x - 4)$
Multiplying by $4$ in RHS (Right Hand Side), we get,
$\Rightarrow$$y - 3 = 4x - 16$
Now, we will shift the terms such that all the constants are on the left-hand side and all the variables on the right-hand side.
$\Rightarrow$$y - 4x = 3 - 16$
Subtracting on RHS,
$\Rightarrow$$y - 4x = - 13$
We need it to be $\operatorname{Ax} + By = C$ , where $\operatorname{A} $ is positive, so we will rearrange and multiply the equation by $ - 1$,
$\Rightarrow$$ - 4x + y = - 13$
Multiply by $ - 1$ on both sides, hence we get,
$\Rightarrow$$4x - y = 13$

Hence, the final equation is $4x - y = 13$.

Note: Suppose we have to find the equation of a line passing through the point $A(1,4)$ and perpendicular to the line joining points \[(2,5)\] , $(2,5)$ and $(4,7)$ . Let us find the slope of the line.
Let the given points be $A(1,4)$ , $B(2,5)$ , $C(4,7)$
Slope of points $BC = \dfrac{{7 - 5}}{{4 - 2}} = \dfrac{2}{2} = 1$
Let, $m = - 1$ ,
The required line also passes through the point $A(1,4)$
The equation is,
$y - {y_1} = m(x - {x_1})$
$y - 4 = - 1(x - 1)$
$y - 4 = - x + 1$
Where the final equation is, $x + y - 5 = 0$ .