Which line is perpendicular to 4x+8y+9=0 and passes through (0,4)

To find the line perpendicular to 4x+8y+9 = 0 through the
point (0,4).

Solution:

A line
through (x1,y1) with a slope m is given by:

y =
m(x-x1)+y1.

Any line through the point (0,4) with a slope m
is given by:

y = m(x-0)+4 or

y
= mx+4. We shall determine by the fact that this line is perpendicular to 4x+8y+ 9 = 0
or

 8y = -4x-9 or

y =
-(1/2)x+9/8, which is in slope intercept form.Therefore the slope (of 4x+8y+9)  is
-1/2.

The two lines y = mx+4 and y = -1/2x-9/8 are
perpendicular only if the product of their slope is
-1.

m*(1/2) = -1.

m =
-2.

So the line y = mx+4 becomes y = 2x+4.
Or

y-2x-4 is the line through (0,4) perpendicular to
4x+8y+9 = 0 is

y -2x-4 = 0 or 2x-y+4 =
0