Question

find the equation of line perpendicular to 5x-3y+1=0 and passing through 4,-3

Answer 1

Answer:

Heya User,

Given Eqn. --> 5x + 3y - 1 = 0

==> y = 1/3[1 - 5x] = -5/3 x + 1/3

--> Slope = -5/3...

---> Slope of perpendicular line = -1/ ( -5/3) = 3/5

==> The eqn. of line is --> [ y - (-4) ] = 3/5 [ x - 3 ]

==> Req.d Eqn. = 5y + 20 = 3x - 9  

==> Req.d Eqn. = 3x - 5y - 29 = 0 <-- Ans..

mark me brilliant  

Answer 2

Answer:

5x-3y+1=0

Step-by-step explanation:

-3y=-5x-1

-y=-5/3x-1/3

the slope of the line is 5/3.

let m stands for the slope perpendicular line then ,5/3m=-1

m=-3/5