# What is the equation of the line that passes through (-3,7) and parallel to the line 2y-4x +2 = 0

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### 3 Answers

Given the point (-3,7) passes through the line.

Then we know that:

y-y1 = m(x-x1)

==> y-7 = m(x+3)

Now we will determine the slope.

Since the line is parallel to the line 2y-4x +2 = 0, then the slopes are equal.

==> We will rewrite into the slope form.

==> y= 2x -1

==> Then the slope is m= 2

==> y-7= 2(x+3)

==> y-7 = 2x + 6

==> y= 2x + 13

**Then the equation of the line is y= 2x+13.**

The slope of two parallel lines is the same.

The equation of the line 2y - 4x + 2 = 0 can be written as

2y = 4x - 2

=> y = 2x - 1

this gives the slope of the line as 2

The line we have to find is parallel to this line and passes through (-3, 7)

The equation of the line would be (y - 7)/(x + 3) = 2

=> y - 7 = 2x + 6

=> y - 2x - 13 = 0

**The required equation of the line is y - 2x - 13 = 0**

Any parallel line to ax+bx+c= 0 is of the form ax+bx+d = 0. The only change is a constant term in the || line.

So 2x-4x+2 has the || line of the form 2y-4x+k = 0. Since 2x-4y+k passes through the point (-3,7), we get 2*(7)-4(-3)+k = 0. So k = 26. Substitute k = 26 in 2y-4x+k = 0. We get 2y -4x+26 = 0 which could be reduced to **y-2x+13= 0 **by dividing by 2.