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  4. The equations of the sides AB , BC and CA of a triangle A B C are 2 x+y=0, x+p y=15 a and x-y=3 respectively. If its orthocentre is (2, a),-(1/2) < a < 2, then p is equal to

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Q. The equations of the sides $AB , BC$ and $CA$ of a triangle $A B C$ are $2 x+y=0, x+p y=15 a$ and $x-y=3$ respectively. If its orthocentre is $(2, a),-\frac{1}{2} < a < 2$, then $p$ is equal to

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Solution:

Coordinates of $A (1,-2), B \left(\frac{15 a }{1-2 p }, \frac{-30 a }{1-2 p }\right)$ and orthocentre $H (2$, a)
Slope of $AH = p$
$a+2=p .....$(1)
Slope of $BH =-1$
$31 a -2 ab =15 a +4 p -2 .....$(2)
From (1) and (2)
$a =1 \& p =3$