Determine the equation of the line that is perpendicular to the line 5x-4y+3=0 and it passes through the point (-1,2) .

We'll  write the equation in the standard
form:

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

We'll use the property of 2 perpendicular lines, that
is:the product between the slopes of 2 perpendicular lines is
:-1

We'll note the 2 slopes as m1 and
m2.

m1*m2=-1

We could find m1
from the given equation of the line which is perpendicular to the one with the unknown
equation.

The equation is
5x-4y+3=0

We'll put the equation into the standard
form:

4y=5x+3

We'll divide by
4 both sides:

y=(5/4)x +3/4 =>
m1=5/4

(5/4)*m2=-1

m2=-4/5

The
equation of a line which passes throuh a given point A(-1,2), and it has the slope
m2 is:

(y-yA)=m(x-xA)

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

We'll
remove the brackets and we'll get:

4x + 5y -10+4 =
0

4x + 5y - 6 =
0