Question 186620
The given line is  3x – 5y = 20 ; adding  5y – 20 on both sides we get

3x – 5y + 5y – 20 = 20 + 5y – 20
3x – 20 = 5y. 		interchanging the sides we get
5y = 3x – 20			Dividing both sides by 5 we get
y  = (3x – 20)/5 = (3/5) x – 4 comparing with y = mx + b we have (3/5) as m

means the slope of this line is 3/5 . A line perpendicular to this line must have its slops as the negative reciprocal of this line. So the slope of the new line = - (5/3) . Let the new line be  
y = (-5/3) x + b    -----  equation 1
Given that the new line passes through ( - 1, 0) . so when its x = -1, y =0 . Let us plug-in these values into equation 1
0 = (-5/3) (-1) + b
That is  0 = (5/3) + b
That is b = ( - 5/3)
Pluging in the value of b in equation 1 we get  y = - ( 5/3) x  - (5/3) 
Which is the required equation