Respuesta :
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
[tex]y = \frac{-4}{5}x + \frac{40}{5}\\\\y = \frac{-4}{5}x + 8[/tex]
On comparing the above equation with eqn 1,
[tex]m = \frac{-4}{5}[/tex]
We know that product of slope of line and slope of line perpendicular to given line is equal to -1
Therefore,
[tex]\frac{-4}{5} \times \text{ slope of line perpendicular to it } = -1\\\\\text{ slope of line perpendicular to it } = \frac{5}{4}[/tex]
Now find the equation of line with slope 5/4 and passing through (-10, -3)
Substitute [tex]m = \frac{5}{4}[/tex] and (x, y) = (-10, -3) in eqn 1
[tex]-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}[/tex]
Substitute [tex]c = \frac{19}{2}[/tex] and [tex]m = \frac{5}{4}[/tex] in eqn 1
[tex]y = \frac{5}{4}x + \frac{19}{2}[/tex]
The standard form of an equation is Ax + By = C
Therefore,
[tex]\frac{5}{4}x - y = -\frac{19}{2}\\\\\frac{5x - 4y}{4} = \frac{-19}{2}\\\\5x - 4y = -38[/tex]
Thus the equation of line in standard form is found
