Question

explain and i will mark u brainlist​

Answer 1

${\huge{\boxed{\mathcal{\orange{Answer}}}}}$

Line AB will be farthest from origin if OP is right angled to the line drawn.

$\blue{\bold{\underline{\underline{Hence,}}}}$

$\sf \green{mOP = \frac{1}{3} \orange\implies mAB = - 3}$

Thus equation of AB is

$\sf \red{(y−1) =−3(x−3)}$

$\sf \pink \implies \purple{a = ( \frac{10}{3} ,0) \: b = (0,10)}$

$\sf \green \implies \blue{a (\triangle \:OAB) = \frac{1}{2}(OA)(OB)} \\ \blue{ = {\frac{1}{2} \times \frac{10}{3} \times 10}}$

$\sf \red\implies\orange {= \frac{100}{6} = \frac{50}{2} sq.units}$

Answer 2

$\orange\bigstar\huge{\colorbox{cyan}{\boxed{\underline{\mathbb{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}}}$

Line AB will be farthest from origin if OP is right angled to the line drawn.

$\red\bigstar\green{\bold{\underline{\underline{Hence,}}}}$

$\sf \green{mOP = \frac{1}{3} \orange\implies mAB = - 3}$

Thus equation of AB is

$\sf \red{(y−1) =−3(x−3)}$

$\sf \pink \implies \purple{a = ( \frac{10}{3} ,0) \: b = (0,10)}$

$\begin{gathered} \sf \green \implies \blue{a (\triangle \:OAB) = \frac{1}{2}(OA)(OB)} \\ \blue{ = {\frac{1}{2} \times \frac{10}{3} \times 10}}\end{gathered}$

$\sf \red\implies\orange {= \frac{100}{6} = \frac{50}{2} sq.units}$