JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Straight Lines

The equations of the sides AB, BC and CA of a triangle ABC 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 ______.

Answer: 3

Solution

Key concept: Orthocentre property — altitude from a vertex is perpendicular to the opposite side.

Find coordinates:

$A$ is intersection of $2 x + y = 0$ and $x - y = 3$ : $A (1, - 2)$
$B$ is intersection of $2 x + y = 0$ and $x + p y = 15 a$ : $B (\frac{15 a}{1 - 2 p}, \frac{- 30 a}{1 - 2 p})$
Orthocentre $H (2, a)$

Condition 1: AH ⊥ BC (slope of BC is

- 1/ p

)
Slope of AH =

\frac{a - ( - 2 )}{2 - 1} = a + 2

Since AH ⊥ BC:

(a + 2) \cdot (- 1/ p) = - 1

a + 2 = p ...(1)

Condition 2: BH ⊥ AC (slope of AC is 1)
Slope of BH =

- 1

\frac{a - \frac{- 30 a}{1 - 2 p}}{2 - \frac{15 a}{1 - 2 p}} = - 1

Simplifying:

31 a - 2 a p = 15 a + 4 p - 2

...(2)

Solving (1) and (2):

a = 1, p = 3

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Integer answer