Write an equation in standard form for the line that passes through (5, -2) and is perpendicular to 3x-4y=12

The equation in standard form for the line that passes through (5, -2) and is perpendicular to 3x - 4y = 12 is 4x + 3y = 14

Given that line that passes through (5, -2) and is perpendicular to 3x - 4y = 12

We have to find the equation of line

The slope intercept form is given as:

y = mx + c  ------ eqn 1

Where "m" is the slope of line and "c" is the y-intercept

Let us first find the slope of line

Given equation of line is 3x - 4y = 12

On rearranging the above equation to slope intercept form,

3x - 4y = 12

4y = 3x - 12

On comparing the above equation with slope intercept form,

Thus the slope of line is

We know that product of slope of given line and slope of line perpendicular to given line is always -1

slope of given line x slope of line perpendicular to given line = -1

Now we have to find the equation of line with slope and passes through (5, -2)

Substitute and (x, y) = (5, -2) in eqn 1

The required equation of line is:

Now substitute and

The standard form of an equation is Ax + By = C

x and y are variables and A, B, and C are integers

Rewriting the above equation,

3y = -4x + 14

4x + 3y = 14

Thus the equation of line in standard form is found out

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50