The equations of the sides AB, BC and CA of a triangle ABC are 2x + y = 0, x + py = 15a and x – y = 3, respectively. If its orthocentre is
\((2, a),−\frac{1}{2}<a<2 \)
then p is equal to _______.
Slope of AH \(=\frac{a+2}{1}\)
Slope of BC\(=−\frac{1}{p}\)
∴ p = a + 2 …(i)
Coordinate of C \(=(\frac{18p−30}{p+1},\frac{15p−33}{p+1})\)
Slope of HC \(=\frac{\frac{15p−33}{p+1}−a}{\frac{18p−30}{p+1}−2}\)
\(=\frac{15p−33−(p−2)(p+1)}{18p−30−2p−2}\)
\(=\frac{16p−p^2−31}{16p−32}\)
\(∵ \frac{16p−p^2−31}{16p−32}×−2=−1\)
∴ p2 – 8p + 15 = 0
∴ p = 3 or 5
But if p = 5 then a = 3 not acceptable
∴ p = 3
So, the answer is 3.
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