Two vertices of a triangle are $(1, 3)$ and $(4, 7)$ . The orthocentre lies on the $x + y = 3$ . The locus of the third vertex is

x^{2} - 2 x y + 2 y^{2} - 3 x - 4 y + 36 = 0

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2 x^{2} - 4 x y + 3 y^{2} 4 x - y + 42 = 0

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4 y^{2} - x y - 3 x^{2} + 2 x - 24 y + 40 = 0

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x^{2} - 4 x y + 3 y^{2} - 2 x - y + 40 = 0

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Solution

The correct option is C $4 y^{2} - x y - 3 x^{2} + 2 x - 24 y + 40 = 0$
$O = (h, 3 - h)$
$m_{B O} = \frac{3 - h - 3}{h - 1} = \frac{- h}{h - 1}$
$m_{C O} = \frac{7 - 3 + h}{4 - h} = \frac{4 + h}{4 - h}$
$A P ⊥ C P$
$m_{A P} = \frac{h - 4}{h + 4}$
$A m ⊥ B m$
$m_{A m} = \frac{h - 1}{h}$
Now
$A p \Rightarrow y - 3 = \frac{h - 4}{h + 4} (x - 1)$
$h (y - 3) + 4 (y - 3) = h (x - 1) - 4 (x - 1)$
$h (y - x - 2) = - 4 y + 12 - 4 x + 4$
$h (y - x - 2) = - (4 y + 4 x - 16)$ ……….. $(1)$
Now
$a m \Rightarrow y - 7 = \frac{h - 1}{h} (x - 4)$
$h (y - 7) = h (x - 4) - (x - 4)$
$h (y - x - 3) = - (x - 4)$ ………… $(2)$
Equation $(1) \div (2)$
$\frac{h (y - x - 3)}{h (y - x - 2)} = \frac{x - 4}{4 y + 4 x - 16}$
$(y - x - 3) (4 y + 4 x - 16) = (x - 4) (y - x - 2)$
$4 y^{2} + 4 x y - 16 y - 4 x y - 4 x^{2} + 16 x - 12 y - 12 x + 48 = x y - 4 y - x^{2} + 4 x - 2 x + 8$
$4 y^{2} - 3 x^{2} - x y + 2 x - 24 y + 40 = 0$ .

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