A line passing through the point of intersect | vedclass

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(B) The given lines are $x+y=4$ and $x-y=2$. Solving these equations, we add them to get $2x=6$, so $x=3$. Substituting $x=3$ into $x+y=4$, we get $y=

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$\frac{32}{9}$

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(B) The given lines are $x+y=4$ and $x-y=2$. Solving these equations, we add them to get $2x=6$, so $x=3$. Substituting $x=3$ into $x+y=4$, we get $y=1$. The point of intersection is $(3, 1)$. The slope of the line is $m = 	an(	an^{-1}(3/4)) = 3/4$. The equation of the line passing through $(3, 1)$ with slope $3/4$ is $(y-1) = \frac{3}{4}(x-3)$, which simplifies to $4y-4 = 3x-9$, or $3x-4y=5$. Thus, $y = \frac{3x-5}{4}$. Substituting this into the parabola equation $y^{2}=4(x-3)$, we get $\left(\frac{3x-5}{4}ight)^{2} = 4(x-3)$. $\frac{9x^{2}-30x+25}{16} = 4x-12$. $9x^{2}-30x+25 = 64x-192$. $9x^{2}-94x+217 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$, we have $x = \frac{94 \pm \sqrt{8836 - 7812}}{18} = \frac{94 \pm \sqrt{1024}}{18} = \frac{94 \pm 32}{18}$. So, $x_{1} = \frac{126}{18} = 7$ and $x_{2} = \frac{62}{18} = \frac{31}{9}$. Therefore, $|x_{1}-x_{2}| = |7 - \frac{31}{9}| = |\frac{63-31}{9}| = \frac{32}{9}$.

$A$ 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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