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...

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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

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