The equation of line in standard form is 5x - 4y = -38
<em><u>Solution:</u></em>
Given that we have to find the equation of line perpendicular to 4x + 5y = 40 that includes the point (-10, -3)
<em><u>The equation of line in slope intercept form is given as:</u></em>
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](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B-4%7D%7B5%7Dx%20%2B%20%5Cfrac%7B40%7D%7B5%7D%5C%5C%5C%5Cy%20%3D%20%5Cfrac%7B-4%7D%7B5%7Dx%20%2B%208)
On comparing the above equation with eqn 1,
![m = \frac{-4}{5}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B-4%7D%7B5%7D)
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}](https://tex.z-dn.net/?f=%5Cfrac%7B-4%7D%7B5%7D%20%5Ctimes%20%5Ctext%7B%20slope%20of%20line%20perpendicular%20to%20it%20%7D%20%3D%20-1%5C%5C%5C%5C%5Ctext%7B%20slope%20of%20line%20perpendicular%20to%20it%20%7D%20%3D%20%5Cfrac%7B5%7D%7B4%7D)
Now find the equation of line with slope 5/4 and passing through (-10, -3)
Substitute
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}](https://tex.z-dn.net/?f=-3%20%3D%20%5Cfrac%7B5%7D%7B4%7D%28-10%29%20%2B%20c%5C%5C%5C%5C-3%20%3D%20%5Cfrac%7B-25%7D%7B2%7D%20%2B%20c%5C%5C%5C%5Cc%20%3D%20-3%20%2B%20%5Cfrac%7B25%7D%7B2%7D%5C%5C%5C%5Cc%20%3D%20%5Cfrac%7B-6%20%2B%2025%7D%7B2%7D%5C%5C%5C%5Cc%20%3D%20%5Cfrac%7B19%7D%7B2%7D)
Substitute
and
in eqn 1
![y = \frac{5}{4}x + \frac{19}{2}](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B5%7D%7B4%7Dx%20%2B%20%5Cfrac%7B19%7D%7B2%7D)
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](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B4%7Dx%20-%20y%20%3D%20-%5Cfrac%7B19%7D%7B2%7D%5C%5C%5C%5C%5Cfrac%7B5x%20-%204y%7D%7B4%7D%20%3D%20%5Cfrac%7B-19%7D%7B2%7D%5C%5C%5C%5C5x%20-%204y%20%3D%20-38)
Thus the equation of line in standard form is found