Find the standard form of equation of a line perpendicular to 4x plus 5y equals 40 that includes the point -10, -3

The equation of line in standard form is 5x - 4y = -38

Solution:

Given that we have to find the equation of line perpendicular to 4x + 5y = 40 that includes the point (-10, -3)

The equation of line in slope intercept form is given as:

y = mx + c ------ eqn 1

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

Given equation of line is:

4x + 5y = 40

Rearranging to slope intercept form, we get

5y = -4x + 40

On comparing the above equation with eqn 1,

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

Therefore,

Now find the equation of line with slope 5/4 and passing through (-10, -3)

Substitute and (x, y) = (-10, -3) in eqn 1

Substitute and in eqn 1

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

Therefore,

Thus the equation of line in standard form is found

first one is a: renaming a fraction. second one is c: de-c

step-by-step explanation:

my explanation is that these were the answers on edg : )

answer: i'm not sure

step-by-step explanation: