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fredd [130]
3 years ago
5

What is the equation of the line that passes through (4, -1) and (-2, 3)?

Mathematics
1 answer:
Mrac [35]3 years ago
7 0

Answer:

2x + 3y - 5 = 0

Step-by-step explanation:

(4 , -1)  & (-2,3)

Slope =\frac{y_{2}-y_{1}}{x_{2}-x_1}}\\\\=\frac{3-[-1]}{-2-4}\\\\=\frac{3+1}{-6}\\\\=\frac{4}{-6}\\\\=\frac{-2}{3}

m = -2/3;   (4 , -1)

y - y1 = m(x - x1)

y - [-1]= \frac{-2}{3}(x-4)\\\\y + 1 =\frac{-2}{3}x -4*\frac{-2}{3}\\\\y + 1 =\frac{-2}{3}x+\frac{8}{3}

Multiply the equation by 3

3y + 3 = -2x + 8

3y = -2x + 8 - 3

3y = -2x + 5

2x + 3y - 5 = 0

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pashok25 [27]

For this case we have to:

\frac {a} {b} = c

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b: It is the Divider

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\frac {5} {x} = y

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ANswer:

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iris [78.8K]

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