# Determine the equation of a line which passes through a point (2,-3) and is parallel to another line 3x-6y+7=0.

*print*Print*list*Cite

The equation of the line is:

y-y1 = m(x-x1) Where (x1,y1) is any point passes through the line, and m is the slope.

We have the point (2,-3) passes through the line. Then:

y-(-3) = m(x-2)

y+3 = m(x-2)

Now we know that the line is parallel to the line 3x-6y +7=0

Then the slope is the same:

==> 3x -6y + 7=0

==> 6y = 3x + 7

==> y= 3/6 x + 7/6

==> y= (1/2)x + 7/6

Then the slope m= 1/2

Now substitute in the line equation:

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

**y= (1/2)x -4**

A line parallel to ax+by+c = o is of the form ax+by+k = 0.

Threfore 3x-6y+ k is a line parallel to 3x-6y+7.

3x-6y+k passes through (2,-3) . So (2,-3) sholud satisfy 3x-6y+k = 0.So,

3(2)-6(-3)+k = 0,

6+18+k = 0.

k = -24'

So the required line is 3x-6y -24 = 0

We know that 2 lines are parallel if the values of their slopes are equal.

The given equation is 3x-6y+7=0 and we have to write in the standard from, which is y = mx+n.

We'll isolate y to the left side:

-6y = -3x-7

We'll divide by -6:

y = (1/2)x + 7/6

So, the slope can be easily determined as m1 = 1/2.

That means that the second slope has the same value:

m2 = 1/2

The line which passes through the point (2,-3) and has the slope m2 = 1/2 has the equation:

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

We'll open the brackets:

y+3 = x/2 -1

We'll add -3 both sides:

y = x/2 -1 - 3

The equation of the parallel line, which passes through (2,-3) is:

**y = x/2 - 4**