finding the equation of a straight line

**johnsy123** · February 15th 2010, 12:02 AM

i'm trying to find the equation of a straight line in y-mx+c from that is parallel to the line with the equation $3x+6y=8, passing through(2,2)$

**Prove It** · February 15th 2010, 12:33 AM

Originally Posted by johnsy123

i'm trying to find the equation of a straight line in y-mx+c from that is parallel to the line with the equation $3x+6y=8, passing through(2,2)$

Write the parallel line you are given in $y = mx + c$ form.

$3x + 6y = 8$

$6y = -3x + 8$

$y = -\frac{1}{2}x + \frac{4}{3}$ .

Parallel lines have the same gradient. The gradient of the parallel line is $-\frac{1}{2}$ , so the line you are trying to find has gradient $-\frac{1}{2}$ .

The line you are trying to find has gradient $m = -\frac{1}{2}$ , and you know the point $(x, y) = (2, 2)$ lies on the line.

So $y = mx + c$

$2 = -\frac{1}{2}(2) + c$

$2 = -1 + c$

$3 = c$ .

So the equation of the line is $y = -\frac{1}{2}x + 3$ .

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