Question

was Find the equation of the line perpendicular to 5x-3y+1=0
and passing through the (4,-3)​

Answer 1

Answer:

3x+5y+3=0

Step-by-step explanation:

We know that for two perpendicular lines with slopes m1 and m2,

m1×m2=-1

So the given equation is 5x-3y+1=0

m1 = 5/3

From this, m2 = -3/5

Now, this line passes from (4,-3)

So equation of line is

(y-y1)=m(x-x1)

=> (y-(-3))=-3(x-4)/5

=> y+3 = (12-3x)/5

=> 5y+15=12-3x

=> 3x+5y+3=0