# Find the vertices of the triangle with sides x - 5y + 8 = 0, 4x - y - 6 = 0, and 3x + 4y + 5 = 0.

1
by japosik1234

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by japosik1234

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x - 5y + 8 = 0 ----equation 1

4x - y - 6 = 0 -----equation 2

3x + 4y + 5 = 0 -----equation 3

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From equations 1 and 2

x - 5y + 8 = 0 ----equation 1

4x - y - 6 = 0 ----equation 2

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Multiply equation 1 with 4

[x - 5y + 8 = 0] x4

4x - 20y + 32 = 0 -----equation 1'

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Subtract equation 1' from 2

4x - y - 6 = 0

- 4x - 20y + 32 = 0

19y - 38 = 0

19y = 38

y = 2

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Substitute y=2 to equation 1

x - 5y + 8 = 0

x - 5(2) + 8 = 0

x - 10 + 8 = 0

x - 2 = 0

x = 2

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The first vertex is at (2,2)

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4x - y - 6 = 0 ------equation 2

3x + 4y + 5 = 0 ------equation 3

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From equation 2

4x - y - 6 = 0

y = 4x - 6 ---equation 2'

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Substitute equation 2' to equation 3

3x + 4y + 5 = 0

3x + 4(4x - 6) + 5 = 0

3x + 16x - 24 + 5 = 0

19x - 19 = 0

19x = 19

x = 1

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Substitute x=1 to equation 2'

y = 4x - 6

y = 4(1) - 6

y = 4 - 6

y = -2

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The second vertex is at (1,-2)

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3x + 4y + 5 = 0 ---equation 3

x - 5y + 8 = 0 ----equation 1

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From equation 1

x - 5y + 8 = 0

x = 5y - 8 ------equation 4

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Substitute equation 4 to equation 3

3x + 4y + 5 = 0

3(5y-8) + 4y + 5 = 0

15y - 24 + 4y + 5 = 0

19y - 19 = 0

19y = 19

y = 1

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Substitute y=1 to equation 4

x = 5y - 8

x = 5(1) - 8

x = 5 - 8

x = -3

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The 3rd vertex is at (-3,1)

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