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

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 |=(10 √10/3) (D) area of triangle A O B equals (50/3), where O is origin

Tardigrade Admin · Accepted Answer

m BA =-3 ⇒ text Equation of line BA ,( y -1)=-3( x -3) ⇒ 3 x + y =10 text or ( x /10 / 3)+( y /10)=1 ∴| AB |=√(100/9)+100=(10 √3/3) Also, ar. ( triangle AOB )=(1/2)((10/3))(10)=(50/3)

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

equation of $A B$ is $4 x + y = 13$

equation of $A B$ is $3 x + y = 10$

$∣ A B ∣ = \frac{10 10}{3}$

area of $△ A OB$ equals $\frac{50}{3}$ , where $O$ is origin

$m_{B A} = - 3$
$\Rightarrow Equation of line B A, (y - 1) = - 3 (x - 3)$
$\Rightarrow 3 x + y = 10 or \frac{x}{10/3} + \frac{y}{10} = 1$
$∴ ∣ A B ∣ = \frac{100}{9} + 100 = \frac{10 3}{3}$
Also, ar. $(△ A OB) = \frac{1}{2} (\frac{10}{3}) (10) = \frac{50}{3}$

Q. 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

equation of AB is 4x+y=13

equation of AB is 3x+y=10

∣AB∣=31010​​

area of △AOB equals 350​, where O is origin