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Question
The vertices of a triangle are A(3, 4), B(2, 0), and C(−1, 6). Find the equation of the line containing the midpoints of sides AB and BC
Solution
Let P, Q be the midpoints of sides AB and BC respectively.
Then P ≡ `((3 + 2)/2, (4 + 0)/2) = (5/2, 2)`
and Q ≡ `((2 - 1)/2, (0 + 6)/2) = (1/2, 3)`
∴ equation of the required line, i.e., line PQ is
`(y - 2)/(x - 5/2) = (3 - 2)/(1/2 - 5/2)`
∴ `(2y - 4)/(2x - 5) = 1/(-2)`
∴ – 4y + 8 = 2x – 5
∴ 2x + 4y – 13 = 0.
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