The equations of the sides AB,BC and CA of a | vedclass

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(D) $1$. The circumcentre $P(2, 3)$ is equidistant from the vertices $A, B, C$. The vertex $A$ is the intersection of $2x + y = 0$ and $x - y = 3$. So

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$34 < 	ext{area}(	riangle ABC) < 38$

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(D) $1$. The circumcentre $P(2, 3)$ is equidistant from the vertices $A, B, C$. The vertex $A$ is the intersection of $2x + y = 0$ and $x - y = 3$. Solving these,we get $A(1, -2)$.$2$. The perpendicular bisector of $AB$ passes through $P(2, 3)$ and is perpendicular to $2x + y = 0$. Its slope is $1/2$. Equation: $y - 3 = \frac{1}{2}(x - 2) \Rightarrow x - 2y + 4 = 0$. Intersection of $AB$ $(2x + y = 0)$ and its perpendicular bisector $(x - 2y + 4 = 0)$ gives the midpoint $M$ of $AB$ as $(-4/5, 8/5)$.$3$. Using the midpoint formula,$B = 2M - A = (-8/5 - 1, 16/5 + 2) = (-13/5, 26/5)$.$4$. Since $B$ lies on $x + py = 39$,we have $-13/5 + p(26/5) = 39$ $\Rightarrow -13 + 26p = 195$ $\Rightarrow 26p = 208$ $\Rightarrow p = 8$.$5$. Vertex $C$ is the intersection of $x + 8y = 39$ and $x - y = 3$. Solving these,we get $C(7, 4)$.$6$. $(AC)^2 = (7 - 1)^2 + (4 - (-2))^2 = 6^2 + 6^2 = 36 + 36 = 72$. Since $p = 8$,$9p = 72$. Thus,$(AC)^2 = 9p$ is true.$7$. $(AC)^2 + p^2 = 72 + 8^2 = 72 + 64 = 136$. This is true.$8$. Area of $	riangle ABC$ with vertices $A(1, -2)$,$B(-13/5, 26/5)$,$C(7, 4)$ is $\frac{1}{2} |1(26/5 - 4) - (-2)(-13/5 - 7) + 1(-52/5 - 182/5)| = \frac{1}{2} |6/5 - 2(48/5) - 234/5| = \frac{1}{2} |6/5 - 96/5 - 234/5| = \frac{1}{2} |-324/5| = 162/5 = 32.4$.$9$. $32 $10$. Therefore,option $D$ is $NOT$ true.

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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