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(B) Given equations are $x+y=4$ $(i)$ and $x-y=2$ $(ii)$.Solving $(i)$ and $(ii)$, we get $x=3$ and $y=1$.The line passing through $(3, 1)$ with slope

Vedclass · Accepted Answer

$\frac{32}{9}$

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(B) Given equations are $x+y=4$ $(i)$ and $x-y=2$ $(ii)$.Solving $(i)$ and $(ii)$, we get $x=3$ and $y=1$.The line passing through $(3, 1)$ with slope $m = 	an(	an^{-1}(3/4)) = 3/4$ is given by:$(y-1) = \frac{3}{4}(x-3) \Rightarrow y = \frac{3x-5}{4}$.Substituting this into the parabola equation $y^{2}=4(x-3)$:$\left(\frac{3x-5}{4}ight)^{2} = 4(x-3)$$\frac{9x^{2}-30x+25}{16} = 4x-12$$9x^{2}-30x+25 = 64x-192$$9x^{2}-94x+217 = 0$.For this quadratic equation, $x_{1}+x_{2} = \frac{94}{9}$ and $x_{1}x_{2} = \frac{217}{9}$.Then $|x_{1}-x_{2}| = \sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}$$= \sqrt{\left(\frac{94}{9}ight)^{2} - 4\left(\frac{217}{9}ight)}$$= \sqrt{\frac{8836}{81} - \frac{868}{9}} = \sqrt{\frac{8836-7812}{81}} = \sqrt{\frac{1024}{81}} = \frac{32}{9}$.

$A$ line passing through the point of intersection of $x+y=4$ and $x-y=2$ makes an angle $\tan^{-1}\left(\frac{3}{4}\right)$ with the $X$-axis. It intersects the parabola $y^{2}=4(x-3)$ at points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$, respectively. Then $|x_{1}-x_{2}|$ is equal to

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