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The area of the triangle formed by the line $x + y = 2$ and angle bisectors of the pair of straight lines $x^{2} - y^{2} + 2 y = 1$ is ‘ $a$ ’ and orthocentre of triangle is $(b, c)$ then $4 a + b + c$ is equal to:

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answer is D.

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Detailed Solution

$x^{2} - y^{2} + 2 y - 1 = 0$

$x^{2} - {(y - 1)}^{2} = 0$

$(x + y - 1) (x - y + 1) = 0$

$a n g l e b i \sec t o r$

$\frac{x + y - 1}{\sqrt{2}} = \pm \frac{x - y + 1}{\sqrt{2}}$

$x + y - 1 = x - y + 1$

$a r e a a = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$

$o r t h o c e n t r e (0, 1) \Rightarrow b = 0, c = 1$

$∴ 4 a + b + c = 2 + 0 + 1 = 3$

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