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3 years ago

Equation of a line, in general form, that is perpendicular to 2x-3y + 7 = 0 and passes through the point (-3/4, 1)

Mathematics

1 answer:

TEA [102]3 years ago

4 0

Answer:

12x +8y +1 = 0

Step-by-step explanation:

You can write the equation by swapping the x- and y-coefficients and negating one of them. Then compute the constant that makes the line go through the given point. You can do that like this:

3(x-(-3/4)) +2(y -1) = 0

3x +9/4 +2y -2 = 0 . . . . eliminate parentheses

3x +2y +1/4 = 0 . . . . . . . simplify

To eliminate the fraction, multiply by 4:

12x +8y +1 = 0

_____

<em>Comment on the equation</em>

The line ax+by=0 passes through the origin. Replacing x and y with x-h and y-k, respectively, makes the line pass through the point (h, k). That's what we did above to make the line pass through the given point.

The business of swapping coefficients and negating one causes the slope of the new line to be the negative reciprocal of the slope of the original line. That is what makes the new line perpendicular to the original.

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Step-by-step explanation:

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3 years ago

What is the midpoints of a segment whose endpoints are (-3, 8) and (7, -2) ?

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$M = (\frac{4}{2}, 3)$
$M = (2, 3)$

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