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 = (-4)/(5)x + (40)/(5)\\\\y = (-4)/(5)x + 8`

On comparing the above equation with eqn 1,

`m = (-4)/(5)`

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

Therefore,

`(-4)/(5) * \text{ slope of line perpendicular to it } = -1\\\\\text{ slope of line perpendicular to it } = (5)/(4)`

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

Substitute
`m = (5)/(4)` and (x, y) = (-10, -3) in eqn 1

`-3 = (5)/(4)(-10) + c\\\\-3 = (-25)/(2) + c\\\\c = -3 + (25)/(2)\\\\c = (-6 + 25)/(2)\\\\c = (19)/(2)`

Substitute
`c = (19)/(2)` and
`m = (5)/(4)` in eqn 1

`y = (5)/(4)x + (19)/(2)`

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

Therefore,

`(5)/(4)x - y = -(19)/(2)\\\\(5x - 4y)/(4) = (-19)/(2)\\\\5x - 4y = -38`

Thus the equation of line in standard form is found