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

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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**