Find the equartion of the line passes through (2,-3) and parallel to the line 8x - 2y +3 = 0

Given the point (2,-3) passes through the
line.

We know that the equation of the line is given by
:

y-y1 = m(x-x1) where m is the slope and (x1,y1) is any
point on the line.

We will substitute with the given point
( 2,-3)

==> y- (-3) = m(
x-2)

==> y+ 3 =
m(x-2)

Now we know that the line is parallel to the line
8x-2y+3 = 0

Then the slopes are
equal,

We will rewrite into the slope
form.

==> -2y = -8x
-3

==> y= 4x + 3/2

Then
the slope is 4.

Now we will substitute into the
equation.

==> y+ 3 =
m(x-2)

==> y+3 =
4(x-2)

==> y+3 = 4x -
8

==> y= 4x
-11

==> Then, the equation of the line
is : y-4x + 11 = 0