Which is the equation of the line that is perpendicular to 4x-5y-1=0 and it passes through the point (2,1) ?

We'll put the equation of the line into the standard
form:

y=mx+n, where m is the slope of the line and n is the
y intercept.

We'll put the equation in the standard form so
we could use the property which says that the product between the slopes of 2
perpendicular lines is :-1

Let's suppose that the 2 slopes
are m1 and m2.

m1*m2=-1

We'll
determine m1 from the given equation of the line, that is perpendicular to the one with
the unknown equation.

The equation is
4x-5y-1=0.

We'll isolate -5y to the left side. For this
reason, we'll subtract 4x-1 both sides:

-5y = -4x +
1

We'll multiply by -1:

5y =
4x - 1

We'll divide by 5:

y =
4x/5 - 1/5

The slope m1 =
4/5.

(4/5)*m2=-1

m2=-5/4

We
also know that the line passes through the point (2,1), so the equation of a line which
passes throuh a given point and it has a well known slope
is:

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

(y-1)=(-5/4)*(x-2)

We'll
remove the brackets:

y - 1 + 5x/4 - 5/2 =
0

y + 5x/4 - 7/2 =
0