A straight line through origin O meets the lines 3y = 10 - 4x and...

Let $A\left({r}_{1}cos\theta, {r}_{1}sin\theta\right)$

$A$ Lies on straight line $4x + 3y = 10$

$⇒ 4\left({r}_{1}cos\theta\right) + 3\left({r}_{1}sin\theta\right) = 10$

$⇒ {r}_{1}\left(4cos\theta + 3sin\theta\right) = 10$
$⇒ {r}_{1} = \dfrac{10}{4cos\theta + 3sin\theta}$

Let $B\left(- {r}_{2}cos\theta, - {r}_{2}sin\theta\right)$

$B$ lies on line $8x + 6y + 5 = 0$

$⇒ 8\left(- {r}_{2}cos\theta\right) + 6\left(- {r}_{2}sin\theta\right) + 5 = 0$

$⇒ {r}_{2}\left(8cos\theta + 6sin\theta\right) - 5 = 0$$⇒ {r}_{2} = \dfrac{5}{8cos\theta + 6sin\theta}$
$∴ \dfrac{{r}_{1}}{{r}_{2}} = \dfrac{4}{1}$