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

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

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

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

Given, $PQ=4$PQ=4 units and the distance between the points ({x}_{1},{y}_{1})&({x}_{2},{y}_{2}) is $\sqrt{{({x}_{1}-{x}_{2})}^{2}+{({y}_{1}-{y}_{2})}^{2}}$(x1​−x2​)2+(y1​−y2​)2​

$$\Rightarrow \sqrt{(\alpha -{2)}^{2}+{(7-\alpha -3)}^{2}}=4$$⇒(α−2)2+(7−α−3)2​=4

$$\Rightarrow \sqrt{(\alpha -{2)}^{2}+{(4-\alpha )}^{2}}=4$$⇒(α−2)2+(4−α)2​=4

$$\Rightarrow (\alpha -{2)}^{2}+{(4-\alpha )}^{2}={4}^{2}$$⇒(α−2)2+(4−α)2=42

$$\Rightarrow {\alpha }^{2}-4\alpha +4+16-8\alpha +{\alpha }^{2}=16$$⇒α2−4α+4+16−8α+α2=16

$$\Rightarrow 2{\alpha }^{2}-12\alpha +4=0$$⇒2α2−12α+4=0

$$\Rightarrow {\alpha }^{2}-6\alpha +2=0$$⇒α2−6α+2=0

$$\Rightarrow \alpha =\frac{-(-6)\pm \sqrt{{(-6)}^{2}-4\times 1\times 2}}{2\times 2}$$⇒α=2×2−(−6)±(−6)2−4×1×2​​

$$\Rightarrow \alpha =\frac{6\pm \sqrt{36-8}}{4}$$⇒α=46±36−8​​

$$\Rightarrow \alpha =\frac{6\pm 2\sqrt{7}}{4}$$⇒α=46±27​​

$$\Rightarrow \alpha =3\pm \sqrt{7}$$⇒α=3±7​

The slope of a line joining the points ({x}_{1},{y}_{1})&({x}_{2},{y}_{2}) is $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}.$x2​−x1​y2​−y1​​.

If we take $\alpha =3-\sqrt{7},$α=3−7​, then $Q\equiv (3-\sqrt{7},4+\sqrt{7})$Q≡(3−7​,4+7​)

And, the slope of $PQ=m=\frac{4+\sqrt{7}-3}{3-\sqrt{7}-2}=\frac{1+\sqrt{7}}{1-\sqrt{7}}$PQ=m=3−7​−24+7​−3​=1−7​1+7​​

If we take $\alpha =3+\sqrt{7},$α=3+7​, then $Q\equiv (3+\sqrt{7},4-\sqrt{7})$Q≡(3+7​,4−7​)

And, the slope of $PQ=m=\frac{4-\sqrt{7}-3}{3+\sqrt{7}-2}=\frac{1-\sqrt{7}}{1+\sqrt{7}}.$PQ=m=3+7​−24−7​−3​=1+7​1−7​​.