A straight line L with negative slope passes through the point (9, 4) and cuts the positive coordinate axes at points P and Q respectively.

Linked Question 1

The area of triangle OPQ, when OP + OQ becomes minimum (where O is the origin) is

125

150

(i) Let equation of line L be (y – 4) = m (x – 9)
$∴ P (9 - \frac{4}{m}, 0), Q (0, 4 - 9 m)$
Now,
$OP + OQ = \underset{\begin{array}{l} positiveq + y \\ as m < 0 \end{array}}{\underset{⏟}{(9 - \frac{4}{m})}} + \underset{\begin{array}{l} positiveq + y \\ as m < 0 \end{array}}{\underset{⏟}{(4 - 9 m)}} = 13 - \frac{4}{m} - 9 m$
$∴ OP + OQ \geq 13 + 2 \sqrt{(\frac{- 4}{m}) (- 9 m)} = 25$
⇒ Minimum value of OP + OQ equal 25, when $\frac{- 4}{m} = - 9 m \Rightarrow m^{2} = \frac{4}{9}$
$∴ m = \frac{- 2}{3}$ (As m < 0)
(ii) So, ar( $∆$ OPQ) = $\frac{1}{2}$ (15)(10) = 75
(iii) Let R(h, k). So
$h = \frac{4}{m}$ …… (1) k = 4 – 9m ……(2)
$∴$ Eliminating m locus of R(h, k) is $(\frac{9}{x} + \frac{4}{y} = 1)$

Linked Question 2

Let R be a moving point on xy plane such that OPRQ becomes a rectangle then the locus of R, as L varies is

$\frac{4}{x} + \frac{9}{y} = 1$

$\frac{x}{9} + \frac{4}{y} = \frac{1}{2}$

$\frac{9}{x} + \frac{4}{y} = 1$

$\frac{x}{9} + \frac{4}{y} = 1$

Linked Question 3

The minimum value of OP + OQ, as L varies, where O is the origin is