# write an equation of the line parallel to x+2y=6 through (8,3)

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`x+2y = 6`

`y = -(1/2)x+3`

The gradient of the parallel lines are the same. So the equation of a line parallel to `y = -(1/2)x+3` is in the form of` y = -(1/2)x+C` . But this line goes through point (8,3).

`y = -(1/2)x+C`

`3 = -(1/2)*8+C`

`C = 7`

*So the equation of the line is ;*

*`y = -(1/2)x+7` *

OR

`2y+x = 14`

**Sources:**

x + 2y = 6

Begin by finding the slope of the original line by putting in slope-intercept form (y=mx+b)

y = -1/2x + 3

Your slope of the original line is 1/2

Now use your point-slope formula

Given the points (8,3)

y - 3 = -1/2(x - 8)

y - 3 = -1/2x + 4

**y = 1/2x + 7**

The slope of the line parellel to the line in question has the exact same slope. You plug the points and the slope into either the point slope formula or standard form.