Respuesta :
The equation of line AB in general form is 5x + 3y - 25 = 0
Solution:
Given that we have to find the equation of line AB
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:
3x - 5y + 17 = 0
Rearrange into slope intercept form
5y = 3x + 17
[tex]y = \frac{3x}{5} + \frac{17}{5}[/tex]
On comparing the above equation with eqn 1,
[tex]m = \frac{3}{5}[/tex]
We know that,
Product of slope of line and slope of line perpendicular to given line is equal to -1
[tex]\frac{3}{5} \times \text{ slope of line perpendicular to given line } = -1\\\\\text{ slope of line perpendicular to given line } = \frac{-5}{3}[/tex]
Now find the equation of line AB with slope [tex]\frac{-5}{3}[/tex] and passes through point C(5,0)
[tex]\text{Substitute } m = \frac{-3}{5} \text{ and } (x, y) = (5, 0) \text{ in eqn 1}[/tex]
[tex]0 = (\frac{-5}{3}) \times 5 + c\\\\0 = \frac{-25}{3} + c\\\\0 = \frac{-25+3c}{3}\\\\-25+3c = 0\\\\3c = 25\\\\Divide\ both\ sides\ of\ equation\ by\ 3\\\\c = \frac{25}{3}[/tex]
[tex]\text{Substitute } c = \frac{25}{3} \text{ and } m = \frac{-5}{3} \text{ in eqn 1 }\\\\y = \frac{-5}{3} \times x + \frac{25}{3}\\\\y = \frac{-5x}{3}+\frac{25}{3}[/tex]
Let us write the equation in general form
The standard form of an equation is Ax + By = C
In this kind of equation, x and y are variables and A, B, and C are integers
[tex]y = \frac{-5x}{3} + \frac{25}{3}\\\\3y = -5x + 25\\\\5x +3y -25 = 0[/tex]
Thus equation of line AB in general form is 5x + 3y - 25 = 0
