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Mathematics Question on Coordinate Geometry

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),−12<a<2(2, a),−\frac{1}{2}<a<2
then p is equal to _______.

Answer

Fig.

Slope of AH =a+21=\frac{a+2}{1}
Slope of BC=−1p=−\frac{1}{p}
∴ p = a + 2 …(i)
Coordinate of C =(18p−30p+1,15p−33p+1)=(\frac{18p−30}{p+1},\frac{15p−33}{p+1})
Slope of HC =15p−33p+1−a18p−30p+1−2=\frac{\frac{15p−33}{p+1}−a}{\frac{18p−30}{p+1}−2}
=15p−33−(p−2)(p+1)18p−30−2p−2=\frac{15p−33−(p−2)(p+1)}{18p−30−2p−2}
=16p−p2−3116p−32=\frac{16p−p^2−31}{16p−32}
∵16p−p2−3116p−32×−2=−1∵ \frac{16p−p^2−31}{16p−32}×−2=−1
∴ p 2 – 8 p + 15 = 0
∴ p = 3 or 5
But if p = 5 then a = 3 not acceptable
∴ p = 3
So, the answer is 3.