# Write the equation of the line that passes through (4,-1) that is parallel to the line 2y+6x=14

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Write the equation of the line parallel to the line `2y+6x=14` that passes through the point (4,-1).

The slope of a line given in standard form `Ax+By+C=0` is `m=(-A)/B` , so the slope of the given line `6x+2y-14=0` is `m=(-6)/2=-3` . A line parallel to this line will have the same slope, e.g. m=-3.

We can use the point-slope form to find the equation of the required line: given slope m and point `(x_1,y_1)` , the equation of the line is `y-y_1=m(x-x_1)` .(This comes from the definition of the slope:`m=(y-y_1)/(x-x_1)==>y-y_1=m(x-x_1)` )

The equation will be `y-(-1)=-3(x-4)==> y+1=-3x+12`

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**In standard form, the equation of the line parallel to 2y+6x=14 passing through the point (4,-1) is 3x+y-11=0 or in slope-intercept form y=-3x+11.**

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To find a line parallel to the one given, rewrite the equation in the form y=mx + b:

2y + 6x = 14

y=1/2 (14 - 6x)

y= -3x + 7

The coefficient of the x, which is m in the general form of the equation, gives you the slope, in this case -3. A line parallel to this one will have the same slope, so you can write down the partial equation:

y = -3x + b.

To write the equation of this parallel line in full, we need to know b. We are given no direct information about b, the y-intercept, but we are given the coordinates of another point (4, -1), so we can substitute these x and y-values in:

-1 = -3(4) + b

b = 11

The equation of the line is thus y= -3x + 11.