A line, with the slope greater than one, passes through the point...

Given point $B$ lie on the line $x - y - 2 = 0$, so the point will be $B\left({x}_{1}, {x}_{1} - 2\right)$

Now given distance between $AB$ is $\dfrac{\sqrt{29}}{3}$, so by distance formula,

$\sqrt{{\left({x}_{1} - 4\right)}^{2} + {\left({x}_{1} - 2 - 3\right)}^{2}} = \dfrac{\sqrt{29}}{3}$

Now on squaring on both side we get,

$18{x}_{1}^{2} - 162{x}_{1} + 340 = 0$

On solving we get, ${x}_{1} = \dfrac{51}{9}$ or ${x}_{1} = \dfrac{10}{3}$

So, ${y}_{1} = \dfrac{33}{9}$ or ${y}_{1} = \dfrac{4}{3}$

Now we can see only $x + 2y = 6$ will satisfy the point \left(\dfrac{10}{3}, \dfrac{4}{3}\right)