Question

Among the lines passing through $C (3,1), BA$ is farthest from the origin and cuts $x$-axis and $y$-axis at $A$ and $B$ respectively, then

• A. equation of $AB$ is $4 x + y =13$
• B. equation of $A B$ is $3 x+y=10$
• C. $| AB |=\frac{10 \sqrt{10}}{3}$
• D. area of $\triangle A O B$ equals $\frac{50}{3}$, where $O$ is origin

Answer 1

Answer: (B), (C), (D)

$ m _{ BA }=-3 $
$\Rightarrow \text { Equation of line } BA ,( y -1)=-3( x -3) $
$\Rightarrow 3 x + y =10 \text { or } \frac{ x }{10 / 3}+\frac{ y }{10}=1$
$\therefore| AB |=\sqrt{\frac{100}{9}+100}=\frac{10 \sqrt{3}}{3}$
Also, ar. $(\triangle AOB )=\frac{1}{2}\left(\frac{10}{3}\right)(10)=\frac{50}{3}$