A variable straight line passes through a fixed point (a, b) intersecting the co-ordinates axes at A B. If ' O ' is the origin, then the locus of the centroid of the triangle OAB is :

Question

A variable straight line passes through a fixed point (a, b) intersecting the co-ordinates axes at A B. If ' O ' is the origin, then the locus of the centroid of the triangle OAB is : (A) b x+a y-3 x y=0 (B) b x+a y-2 x y=0 (C) a x+b y-3 x y=0 (D) a x+b y-2 x y=0

Tardigrade Admin · Accepted Answer

equation of line A B y-b=m(x-a) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/4dad150ac9808c4f9a869f87f9eb7ed2-.png" /> ∴ G (( a -( b / m )/3), ( b - am /3)) ⇒ h =( a -( b / m )/3), k =( b - am /3) on eleminating ' m ' we get required locus b h+a k-3 h k=0 ⇒ b x+a y-3 x y=0

Q. A variable straight line passes through a fixed point $(a, b)$ intersecting the co-ordinates axes at $A & B$ . If ' $O$ ' is the origin, then the locus of the centroid of the triangle $O A B$ is :

$b x + a y - 3 x y = 0$

$b x + a y - 2 x y = 0$

$a x + b y - 3 x y = 0$

$a x + b y - 2 x y = 0$

equation of line $A B$
$y - b = m (x - a)$

$∴ G (\frac{a - \frac{b}{m}}{3}, \frac{b - am}{3})$
$\Rightarrow h = \frac{a - \frac{b}{m}}{3}, k = \frac{b - am}{3}$
on eleminating ' $m$ ' we get required locus
$bh + ak - 3 hk = 0 \Rightarrow b x + a y - 3 x y = 0$

Q. A variable straight line passes through a fixed point (a,b) intersecting the co-ordinates axes at A&B. If ' O ' is the origin, then the locus of the centroid of the triangle OAB is :

bx+ay−3xy=0

bx+ay−2xy=0

ax+by−3xy=0

ax+by−2xy=0