Slope of a line passing through P ( 2, 3 ) and intersecting the...

Given, the point $$P\left(2, 3\right)$ on the line $x + y = 7 .$

Let, the point of intersection of the two lines be $Q$

Then, $Q$ can be taken as $Q ≡ ( \alpha, 7 - \alpha )$

Given, $PQ = 4$ units and the distance between the points \left({x}_{1}, {y}_{1}\right) \text{ and } $\left({x}_{2}, {y}_{2}\right)$ is $\sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$

$⇒ \sqrt{( \alpha - {2 )}^{2} + {\left(7 - \alpha - 3\right)}^{2}} = 4$

$⇒ \sqrt{( \alpha - {2 )}^{2} + {\left(4 - \alpha\right)}^{2}} = 4$

$⇒ ( \alpha - {2 )}^{2} + {\left(4 - \alpha\right)}^{2} = {4}^{2}$

$⇒ {\alpha}^{2} - 4\alpha + 4 + 16 - 8\alpha + {\alpha}^{2} = 16$

$⇒ 2{\alpha}^{2} - 12\alpha + 4 = 0$

$⇒ {\alpha}^{2} - 6\alpha + 2 = 0$

$⇒ \alpha = \dfrac{- \left(- 6\right) \pm \sqrt{{\left(- 6\right)}^{2} - 4 \times 1 \times 2}}{2 \times 2}$

$⇒ \alpha = \dfrac{6 \pm \sqrt{36 - 8}}{4}$

$⇒ \alpha = \dfrac{6 \pm 2\sqrt{7}}{4}$

$⇒ \alpha = 3 \pm \sqrt{7}$

The slope of a line joining the points \left({x}_{1}, {y}_{1}\right) \text{ and } $\left({x}_{2}, {y}_{2}\right)$ is $\dfrac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} .$

If we take $\alpha = 3 - \sqrt{7},$ then $Q ≡ \left(3 - \sqrt{7}, 4 + \sqrt{7}\right)$

And, the slope of $PQ = m = \dfrac{4 + \sqrt{7} - 3}{3 - \sqrt{7} - 2} = \dfrac{1 + \sqrt{7}}{1 - \sqrt{7}}$

If we take $\alpha = 3 + \sqrt{7},$ then $Q ≡ \left(3 + \sqrt{7}, 4 - \sqrt{7}\right)$

And, the slope of $PQ = m = \dfrac{4 - \sqrt{7} - 3}{3 + \sqrt{7} - 2} = \dfrac{1 - \sqrt{7}}{1 + \sqrt{7}} .