# Find the equation of the line passes through the point ( 5,-2) and parallel to the line : 2y-x = 5

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First we will write the equation for the line in the standard format:

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

==> y--2 = m(x-5)

==> y+ 2 = m(x-5)

But given the the line 2y-x = 5 is perpendicular to the line.

Then the products of the slopes should be -1.

2y-x = 5

We will rewrite in slope form:

2y= x+ 5

==> y= (1/2)x + 5/2

Then the slpe = 1/2

==> 1/2* m = -1

==> m= -2

==> y+ 2 = -2(x-5)

==> y= -2x + 10 -2

**==> y= -2x +8**

Two lines are parallel if and only if their slopes are equal or if the ratio of their correspondent coefficients are also equal.

We'll put the given equation into the standard form. For this reason, we'll isolate 2y to the left side:

2y = x + 5

We'll divide by 2:

y = x/2 + 5/2

We'll write the standard form of the equation of the parallel line and we'll identify the value of the slope for both lines.

y = mx + n

m1 = 1/2 and m2 = m

The slopes have like values.

m = 1/2

Now, we'll write the equtaion of the line that has the slope m=1/2 and it passes through the point (5,-2).

y - (-2) = (1/2)(x - 5)

y + 2 = x/2 - 5/2

y = x/2 - 5/2 - 2

y = x/2 - 9/2

2y = x - 9

2y - x + 9 = 0

**The equation of the parallel line whose slope is m = 1/2 and it is passing through the point (5,-2) is 2y - x + 9 = 0. **