A line through the variable point $A(k+1, 2k)$ meets the lines $7 x + y - 16 = 0, 5 x - y - 8 = 0, x - 5 y + 8 = 0$ at B. C, D respectively, then AC, AB, AD are in

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Detailed Solution

Given lines are $7 x + y - 16 = 0$ ……… (1)
   $5 x - y - 8 = 0$ ……… (2)
   $x - 5 y + 8 = 0$ ……… (3)
Let the equation of line passing through $A (k + 1, 2 k)$ making an angle $θ$ with the +ve direction of x-axis, be
$\frac{x - (k + 1)}{\cos θ} = \frac{y - 2 k}{\sin θ} = r_{1}, r_{2}, r_{3}$ (if $A B = r_{1}, A C = r_{2}, A D = r_{3}$ )
$B \equiv [(k + 1) + r_{1} \cos θ, 2 k + r_{1} \sin θ]$
$C \equiv [(k + 1) + r_{2} \cos θ, 2 k + r_{2} \sin θ]$
$D \equiv [(k + 1) + r_{3} \cos θ, 2 k + r_{3} \sin θ]$
Points B, C, D satisfying (1), (2) and (3) respectively then
$r_{1} = \frac{9 (1 - k)}{7 \cos θ + \sin θ}$
$r_{2} = \frac{3 (1 - k)}{5 \cos θ - \sin θ}$
and $r_{3} = \frac{9 (1 - k)}{5 \cos θ - \sin θ}$
$\frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{(5 \cos θ - \sin θ)}{3 (1 - k)} + \frac{(5 \sin θ - \cos θ)}{9 (1 - k)}$
$= \frac{15 \cos θ - 3 \sin θ + 5 \sin θ - \cos θ}{9 (1 - k)}$
$= \frac{14 \cos θ + 2 \sin θ}{9 (1 - k)} = \frac{2}{r_{1}}$
Hence    $r_{2}, r_{1}, r_{3}$ are in HP
i.e, AC, AB, AD are in HP