Write an equation of the line that passes through (2, -5) and is parallel to the line 2y = 3x + 10

The equation of the line which passes through (2, -5) and is parallel to the line 2y = 3x + 10 is y = 3/2x - 8

Given the points : (2, -5)

The standard equation given : 2y = 3x + 10

Make y the subject ;

Divide both sides by 2

According to slope - intercept relation : y = bx + c

Since the point are parallel ; then they'll share the same slope

Using the values of X = 2 and y = - 5

-5 = 3/2(2) + c

-5 = 3 + c

c = - 5 - 3

c = - 8

The slope-intercept relation for the point (2,-5) is;

Therefore, the equation of the line is y = 3/2x - 8

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First you get "y" by itself. To do so you divide 2 on both sides.

y = 3/2x + 5

To write an equation of a line that is PARALLEL to this equation, the slopes have to be the SAME. So the slope is 3/2.

You then use the equation:

y = mx + b

SInce you know "m" you plug it in.

y = 3/2x + b

Now you need to find b. To do so you plug in the point (2, -5) into this equation.

-5 = 3/2(2) + b

-5 = 3 + b

-8 = b

Finally you plug in b and you get your new equation.

y = 3/2x - 8

the answer is d

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