The slope of a line passing through P(2, 3) a | vedclass

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(D) Let the slope of the line be $m = 	an 	heta$. The equation of the line passing through $P(2, 3)$ is $x = 2 + r \cos 	heta$ and $y = 3 + r \sin

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$\frac{1 - \sqrt{7}}{1 + \sqrt{7}}$

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(D) Let the slope of the line be $m = 	an 	heta$. The equation of the line passing through $P(2, 3)$ is $x = 2 + r \cos 	heta$ and $y = 3 + r \sin 	heta$,where $r = 4$.Substituting these into $x + y = 7$:$(2 + 4 \cos 	heta) + (3 + 4 \sin 	heta) = 7$$4(\cos 	heta + \sin 	heta) = 2$$\cos 	heta + \sin 	heta = \frac{1}{2}$Squaring both sides:$1 + 2 \sin 	heta \cos 	heta = \frac{1}{4}$$\sin 2	heta = \frac{1}{4} - 1 = -\frac{3}{4}$Using $\sin 2	heta = \frac{2m}{1 + m^2}$:$\frac{2m}{1 + m^2} = -\frac{3}{4}$$8m = -3 - 3m^2 \Rightarrow 3m^2 + 8m + 3 = 0$Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$m = \frac{-8 \pm \sqrt{64 - 36}}{6} = \frac{-8 \pm \sqrt{28}}{6} = \frac{-8 \pm 2\sqrt{7}}{6} = \frac{-4 \pm \sqrt{7}}{3}$Rationalizing $\frac{1 - \sqrt{7}}{1 + \sqrt{7}} = \frac{(1 - \sqrt{7})^2}{1 - 7} = \frac{1 + 7 - 2\sqrt{7}}{-6} = \frac{8 - 2\sqrt{7}}{-6} = \frac{-4 + \sqrt{7}}{3}$.

The slope of a line passing through $P(2, 3)$ and intersecting the line $x + y = 7$ at a distance of $4$ units from $P$ is

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