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Question
Answer the following question:
The vertices of a triangle are A(1, 4), B(2, 3) and C(1, 6). Find equations of the sides.
Solution
The vertices of triangle are A(1, 4), B(2, 3) and C(1, 6).
∴ equation of side AB is
`(y - 4)/(x - 1) = (3 - 4)/(2 - 1) = (-1)/1` = – 1
∴ y – 4 = –x + 1
∴ x + y = 5
Equation of side BC is
`(y - 3)/(x - 2) = (6 - 3)/(1 - 2) = 3/(-1)` = – 3
∴ y – 3 = – 3x + 6
∴ 3x + y = 9
Equation of side AC is
`(y - 4)/(x - 1) = (6 - 4)/(1 - 1) = 2/0`
∴ 0 = 2x – 2
∴ x = 1
Hence, equations of the sides of the triangle are x + y = 5, 3x + y = 9 and x = 1.
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