Write an equation of a line that is perpendicular to 2y = 5/3x + 10 and passes through the point (10, -3)

Hello there!

I’m assuming you mean y=(5/3)x+10 correct me if there’s supposed to be a 2 in there.

First of all, to get a slope perpendicular to 5/3, flip the fraction and multiply it by negative 1.
-3/5

So now we have the line:
y=(-3/5)x+b

Now plug the point (10,-3) into the equation.
-3=(-3/5)10+b
-3=-30/5+b
-3=-6+b
3=b

Now plug the 3 in for b into the equation.
y=(-3/5)x+3

I hope this helps!
Best wishes :)

-HuronGirl