# Find the equation of the line passes through ( -3, 4) and parallel to the line 2y-4x + 2 = 0

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Given the line passes through the point ( -3, 4).

Then, we will write the equation of the line:

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

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

Now we need to determine the slope m.

But, we know that the line is parallel to the line 2y-4x + 2 = 0

Then the slopes are equal.

We will write the equation into the slope form.

==> 2y= 4x -2

==> y= 2x -1

Then the slope is m = 2

We will substitute into the equation of the line.

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

==> y-4 = 2x + 6

**==> y= 2x + 10**

If 2 lines are parallel, their slopes have to be equal.

We'll put the given equation of the line in the point slope form.

y = mx + n

For this reason, we'll have to isolate y to the left side and we'll add 4x and subtract 2 both sides:

2y = 4x - 2

We'll divide by 2:

y = 2x - 1

Comparing, we'll get the slope of the first line: m1 = 2

The slope of the parallel line is m2 = 2.

The line is passing through the point (-3 , 4). The equation of the parallel line is:

y - 4 = 2(x + 3)

We'll add 4:

y = 2(x + 3) + 4

We'll remove the brackets:

y = 2x + 6 + 4

We'll combine like terms:

y = 2x + 10

**The equation of the parallel line is: **

**y = 2x + 10**

Any line parallel to the given line ax+by +c = 0 is of the form ax+by+d = 0. If the line ax+by+d passes through a given point (x1, x2), then the equation of the parallel line is a(x-x1)+b(y-y1) = 0.

So the given line is 2y-4x+2 = 0, or 4x-2y-2 = 0.

So any line parallel to 4x-2y-2 = 0 and passing through (-3, 4) is 4(x- (-3))-2(y-4) = 0.

=> 4x+12-2y+8 = 0

=> 4x-2y+20 = 0.

Therefore 4x-2y+20=0 is the line which is parallel to 2y-4x+2 = 0 and passes through the point (-3, 4).