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प्रश्न
Find the equations of the medians of a triangle, the equations of whose sides are:
3x + 2y + 6 = 0, 2x − 5y + 4 = 0 and x − 3y − 6 = 0
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उत्तर
The given equations are as follows:
3x + 2y + 6 = 0 ... (1)
2x − 5y + 4 = 0 ... (2)
x − 3y − 6 = 0 ... (3)
In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.
Solving (1) and (2):
x = −2, y = 0
Thus, AB and BC intersect at B (−2, 0).
Solving (1) and (3):
x = \[- \frac{6}{11}\] , y = \[- \frac{24}{11}\]
Thus, AB and CA intersect at \[A \left( - \frac{6}{11}, - \frac{24}{11} \right)\].
Similarly, solving (2) and (3):
x = −42, y = −16
Thus, BC and CA intersect at C (−42, −16).
Let D, E and F be the midpoints the sides BC, CA and AB, respectively.Then,
Then, we have:
\[D = \left( \frac{- 2 - 42}{2}, \frac{0 - 16}{2} \right) = \left( - 22, - 8 \right)\]
\[E = \left( \frac{- \frac{6}{11} - 42}{2}, \frac{- \frac{24}{11} - 16}{2} \right) = \left( \frac{- 234}{11}, - \frac{100}{11} \right)\]
\[F = \left( \frac{- \frac{6}{11} - 2}{2}, \frac{- \frac{24}{11} + 0}{2} \right) = \left( - \frac{14}{11}, - \frac{12}{11} \right)\]
Now, the equation of median AD is
\[y + \frac{24}{11} = \frac{- 8 + \frac{24}{11}}{- 22 + \frac{6}{11}}\left( x + \frac{6}{11} \right)\]
\[ \Rightarrow 16x - 59y - 120 = 0\]
The equation of median BE is \[y - 0 = \frac{- \frac{100}{11} - 0}{- \frac{234}{11} + 2}\left( x + 2 \right)\]
\[ \Rightarrow 25x - 53y + 50 = 0\]
And, the equation of median CF is
\[y + 16 = \frac{- \frac{12}{11} + 16}{- \frac{14}{11} + 42}\left( x + 42 \right)\]
\[ \Rightarrow 41x - 112y - 70 = 0\]
