A line, with the slope greater than one, passes through the point A(4, 3) and intersects the line x – y – 2 = 0 at the point B. If the length of the line segment AB is $\frac{\sqrt{29}}{3}$ , then B also lies on the line :

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$2 x + y = 9$

$x + 2 y = 6$

$3 x - 2 y = 7$

$2 x - 3 y = 3$

answer is C.

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Detailed Solution

Let $B (x_{1}, x_{1} - 2)$

$\sqrt{{(x_{1} - 4)}^{2} + {(x_{1} - 2 - 3)}^{2}} = \frac{\sqrt{29}}{3}$

Squaring on both side

$\begin{array}{l} 18 {x_{1}}^{2} - 162 x_{1} + 340 = 0 \\ x_{1} = \frac{51}{9} or x_{1} = \frac{10}{3} \\ y_{1} = \frac{33}{9} or y_{1} = \frac{4}{3} \end{array}$

Option (C) will satisfy $(\frac{10}{3}, \frac{4}{3})$

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