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