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)

Question

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 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Coordinates of A (1,-2), B ((15 a /1-2 p ), (-30 a /1-2 p )) 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

Q. The equations of the sides $A B, BC$ and $C A$ of a triangle $A BC$ 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

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

Q. 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),−21​<a<2, then p is equal to

Coordinates of A(1,−2),B(1−2p15a​,1−2p−30a​) and orthocentre H(2, a) Slope of AH=p a+2=p.....(1) Slope of BH=−1 31a−2ab=15a+4p−2.....(2) From (1) and (2) a=1&p=3

Q. The equations of the sides $A B, BC$ and $C A$ of a triangle $A BC$ 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

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