How do you find in standard form a line that is parallel to $- 2 x - 3 y = - 2$ $-2x-3y = -2$ and passing through (2, -2)?

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How do you find in standard form a line that is parallel to $- 2 x - 3 y = - 2$ $-2x-3y = -2$ and passing through (2, -2)?

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Answer 1 · 2018-03-23T16:04:32

$- 2 x - 3 y = 2$ $-2x-3y=2$

Explanation:

First, we need to find the slope of the line that it is parallel to, as this line will have the same slope.
$- 2 x - 3 y = - 2$ $-2x-3y=-2$
$3 y = - 2 x - 2$ $3y=-2x-2$
$y = - \frac{2}{3} x - \frac{2}{3}$ $y=-2/3x-2/3$
Therefore, $m = - \frac{2}{3}$ $m=-2/3$
We can use point-slope form at first before we convert it to standard form.
$y - y_{1} = m (x - x_{1})$ $y-y_1=m(x-x_1)$
$y + 2 = - \frac{2}{3} (x - 2)$ $y+2=-2/3(x-2)$
$y + 2 = - \frac{2}{3} x + \frac{4}{3}$ $y+2=-2/3x+4/3$
$3 y + 6 = - 2 x + 4$ $3y+6=-2x+4$ multiply by 3 to get rid of fractions
$- 2 x - 3 y = 2$ $-2x-3y=2$

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How do you find in standard form a line that is parallel to $- 2 x - 3 y = - 2$ $-2x-3y = -2$ and passing through (2, -2)?

1 Answer

Explanation:

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How do you find in standard form a line that is parallel to −2x−3y=−2-2x-3y = -2 and passing through (2, -2)?

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How do you find in standard form a line that is parallel to $- 2 x - 3 y = - 2$ $-2x-3y = -2$ and passing through (2, -2)?