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प्रश्न
Find the equations of the medians of a triangle, the coordinates of whose vertices are (−1, 6), (−3, −9) and (5, −8).
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उत्तर
Let A (−1, 6), B (−3, −9), and C (5, −8) be the coordinates of the given triangle.
Let D, E and F be midpoints of BC, CA and AB, respectively.
So, the coordinates of D, E, and F are:

D = \[\left( \frac{- 3 + 5}{2}, \frac{- 9 - 8}{2} \right)\]
∴ D = \[\left( 1, \frac{- 17}{2} \right)\]
E = \[\left( \frac{- 1 + 5}{2}, \frac{6 - 8}{2} \right)\]
∴ E = (2, −1)
F = \[\left( \frac{- 1 - 3}{2}, \frac{6 - 9}{2} \right)\]
∴ F = \[\left( - 2, - \frac{3}{2} \right)\]
Median AD passes through,
A(−1, 6) and D\[\left( 1, - \frac{17}{2} \right)\]
So, its equation is:
\[y - 6 = \frac{- \frac{17}{2} - 6}{1 + 1}\left( x + 1 \right)\]
⇒ 4y − 24 = −29x − 29
⇒ 29x + 4y + 5 = 0
Median BE passes through B (−3, −9) and E(2, −1)
So, its equation is:
\[y + 9 = \frac{- 1 + 9}{2 + 3}\left( x + 3 \right)\]
⇒ 5y + 45 = 8x + 24
⇒ 8x − 5y − 21 = 0
Median CF passes through,
C (5, −8) and F \[\left( - 2, - \frac{3}{2} \right)\]
So, its equation is:
\[y + 8 = \frac{- \frac{3}{2} + 8}{- 2 - 5}\left( x - 5 \right)\]
⇒ −14y − 112 = 13x − 65
⇒13x + 14y + 47 = 0
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