Question

find the standard form of equation of a line perpendicular to 4X plus 5Y equals 40 that includes the point -10, -3

Answer 1

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

$y = \frac{-4}{5}x + \frac{40}{5}\\\\y = \frac{-4}{5}x + 8$

On comparing the above equation with eqn 1,

$m = \frac{-4}{5}$

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

Therefore,

$\frac{-4}{5} \times \text{ slope of line perpendicular to it } = -1\\\\\text{ slope of line perpendicular to it } = \frac{5}{4}$

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

Substitute $m = \frac{5}{4}$ and (x, y) = (-10, -3) in eqn 1

$-3 = \frac{5}{4}(-10) + c\\\\-3 = \frac{-25}{2} + c\\\\c = -3 + \frac{25}{2}\\\\c = \frac{-6 + 25}{2}\\\\c = \frac{19}{2}$

Substitute $c = \frac{19}{2}$ and $m = \frac{5}{4}$ in eqn 1

$y = \frac{5}{4}x + \frac{19}{2}$

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

Therefore,

$\frac{5}{4}x - y = -\frac{19}{2}\\\\\frac{5x - 4y}{4} = \frac{-19}{2}\\\\5x - 4y = -38$

Thus the equation of line in standard form is found