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Q.

A straight line L with negative slope passes through the point (8, 2) and cuts positive co-ordinate axes at the
points P and Q. Find the minimum value of $O P + O Q$ as L varies where ‘O’ is the origin.

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answer is 18.

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Detailed Solution

Question Image

$θ > 90 \Rightarrow m < 0$

Equation of L is

$y - 2 = m (x - 8)$

$m x - y = \frac{8 m - 2}{- 1}$

$O P = \frac{8 m - 2}{m}, O Q = \frac{8 m - 2}{- 1}$

$Let f (x) = \frac{8 m - 2}{m} - (8 m - 2)$

$8 - \frac{2}{m} - 8 m + 2 = 10 - (8 m + \frac{2}{m})$

$= 10 + (8 (- m) + \frac{2}{(- m)})$

$\geq 10 + 2 \sqrt{8 .2}$

$\geq 10 + 8 = 18$

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A straight line L with negative slope passes through the point (8, 2) and cuts positive co-ordinate axes at thepoints P and Q. Find the minimum value of OP+OQ as L varies where ‘O’ is the origin.