The equations of the sides AB, BC and CA of a triangle ABC are 2x + 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 :

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

34 < area ( $∆$ ABC) < 38

${(AC)}^{2} = 9 p$

32 < area ( $∆$ ABC) < 36

answer is D.

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Detailed Solution

Perpendicular bisector of AB

x + y = 5

Take image of A

$\begin{array}{l} \frac{x - 1}{1} = \frac{y + 2}{1} = \frac{- 2 (- 6)}{2} = 6 \\ (7, 4) \\ 7 + 4 p = 39 \\ p = 8 \\ solving x + 8 y = 39 and y = - 2 x \\ x = \frac{- 39}{15} y = \frac{78}{15} \\ {AC}^{2} = 72 = 9 p \\ {AC}^{2} + p^{2} = 72 + 64 = 136 \\ ΔABC = \frac{1}{2} |\begin{array}{ccc} 1 & - 2 & 1 \\ 7 & 4 & 1 \\ \frac{- 39}{15} & \frac{78}{15} & 1 \end{array}| \\ = \frac{1}{2} [4 - \frac{78}{15} + 2 (7 + \frac{39}{15}) + 7 (\frac{78}{15}) + \frac{4 \times 39}{15}] \\ = \frac{1}{2} [18 + 18 \times \frac{13}{5}] \\ = 9 [\frac{18}{5}] = \frac{162}{5} = 32 .4 \end{array}$