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

Sagot :

Shinalcantaraidn
To find the coordinates of the vertices of the triangle you'll have:
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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