Question 186716
You have given the condition but not the question. I presume you want the equation of a line passing through (4,- 5) and perpendicular to the line 
3y =  - (2/5)x +3

Ans: The condition for a line to be perpendicular to another line is that the product of their slopes must be  - 1 . Means, one slope must be the  negative reciprocal of the other

The given line is 3y = - (2/5) x + 3 . Rewriting it in y = mx+ b form we have

y = - (2/15) x + 1. So its slope = - (2/15) . The required line being perpendicular to this must have a slope =  15/2  and it passes through (4,- 5)

Let it be  y = mx +b                        …. Eqn(1)
. when x = 4, y = -5 since it passes through (4, - 5).  Plugging in these values ( m= 15/2, x = 4, y = - 5)into eqn (1)  we get   - 5 = (15/5) (4)  + b
-5 = 12 + b ;  so b = -5 -12 = -17 . Now we shall substitute the values of m and b in eqn(1)
y = (15/2) x – 17  which is the required equation. We can multiply it by 2 and  make it into the standard form 

2y = 15x – 17 ;    ie. 15x – 2y – 17 = 0  is the required line.




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