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Write an equation for the line passing through (-4,7) and parallel to the line 8x−3y−4=0.

a) 8x−3y−4=0
b) 8x−3y+29=0
c) 8x−3y+20=0
d) 8x−3y−12=0

1 Answer

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Final answer:

The equation for the line passing through (-4,7) and parallel to the line 8x−3y−4=0 is 8x−3y+53=0.

Step-by-step explanation:

To find the equation of a line parallel to another line, we need to determine the slope of the given line.

The equation of the given line is 8x - 3y - 4 = 0.

To find the slope, we can rearrange the equation in slope-intercept form, which is in the form y = mx + b, where m is the slope. So, rearranging the equation, we get:

8x - 3y = 4

-3y = -8x + 4

y = (8/3)x - (4/3)

The slope of the given line is 8/3.

Since the line we want is parallel to the given line, it will have the same slope.

Now, we can use the point-slope form of a line to write the equation of the line passing through (-4, 7) with a slope of 8/3:

y - y1 = m(x - x1)

Substituting the values, we get:

y - 7 = (8/3)(x - (-4))

y - 7 = (8/3)(x + 4)

Multiplying through by 3 to eliminate the fraction, we get:

3y - 21 = 8(x + 4)

3y - 21 = 8x + 32

8x - 3y + 53 = 0

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