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