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MaRussiya [10]
3 years ago
6

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

Mathematics
1 answer:
Zarrin [17]3 years ago
8 0

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

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The coordinates of the point of intersection is (2, 7)

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The first straight line graph from the left, we have;

y-intercept = (0, 5), x-intercept = (-5, 0) therefore, the slope, m = (0 - 5)/(-5 - 0) = 1

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The equation of the graph in slope-intercept form is y - 9 = -1×(x - 0)

Which gives;

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To find the point of intersection, we equate equation (1) to equation (2), which are the y-value function of x, we have;

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From either equation (1) or equation (2), we can find the value of y as follows;

y = -x + 9, where x = 2 gives;

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