The equations of the sides $\mathrm{AB}, \mathrm{BC}$ and CA of a triangle ABC are $2 \mathrm{x}+\mathrm{y}=0, x+\mathrm{py}=39$ and $\mathrm{x}-\mathrm{y}=3$ respectively and $\mathrm{P}(2,3)$ is its circumcentre. Then which of the following is NOT true :

(D) $34<$ area $(\triangle \mathrm{ABC})<38$

(A) $(\mathrm{AC})^{2}=9 \mathrm{p}$

(B) $(\mathrm{AC})^{2}+\mathrm{p}^{2}=136$

(C) $32<$ area $(\triangle \mathrm{ABC})<36$

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Straight Lines

The equations of the sides

AB, BC

and CA of a triangle ABC are

2 x + y = 0, x + py = 39

and

x - y = 3

respectively and

P (2, 3)

is its circumcentre. Then which of the following is NOT true :

Answer: (D)

Vertex

A

is the meeting of

2 x + y = 0

and

x - y = 3

, giving

A (1, - 2)

, for which the circumcentre satisfies

P A^{2} = 26

. The remaining vertices lie one on

2 x + y = 0

and one on

x - y = 3

and must also be at distance

26

from

P (2, 3)

; solving

P B^{2} = 26

along

2 x + y = 0

gives

B (- \frac{13}{5}, \frac{26}{5})

, and

P C^{2} = 26

along

x - y = 3

gives

C (7, 4)

. Substituting

C

into

x + p y = 39

fixes

p = 8

. Then

A C^{2} = (7 - 1)^{2} + (4 + 2)^{2} = 72 = 9 p

, and

A C^{2} + p^{2} = 72 + 64 = 136

, so both of those statements hold. The area is

\frac{1}{2} 1 (\frac{26}{5} - 4) - \frac{13}{5} (4 + 2) + 7 (- 2 - \frac{26}{5}) = \frac{162}{5} = 32.4

, which satisfies

32 < area < 36

but not

34 < area < 38

, so it is the latter inequality that fails to be true.

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option