Question

Find locus of centroid of triangle aob if line ab passes through (3,2) a and b are on coordinate axes

Answer 1

Solution:

Concept used:

The centroid of a triangle having vertices

$({x_1},{y_1}),({x_2},{y_2}), ({x_3},{y_3})\: is\:(\frac{{x_1}+{x_2}+{x_3}}{3},\frac{{y_1}+{y_2}+{y_3}}{3})$

since the line interssect the coordinate axes at A and B,

the coordinates of A and B are (a,0) and (0,b).

The equation fo the given line intercept

form is

$\frac{x}{a}+\frac{y}{b}=1$

since line passes through (3,2),

$\frac{3}{a}+\frac{2}{b}=1\\\\\frac{3b+2a}{ab}=1$

3b+2a=ab..........(1)

Let the centroid of triangle AOB be (h,k).

Then,

$(\frac{{x_1}+{x_2}+{x_3}}{3},\frac{{y_1}+{y_2}+{y_3}}{3})=(h,k)\\\\(\frac{a+0+0}{3},\frac{0+b+0}{3})=(h,k)\\\\(\frac{a}{3},\frac{b}{3})=(h,k)\\\\\frac{a}{3}=h,\:\frac{b}{3}=k$

a=3h, b= 3k

Now (1) becomes

3(3k)+2(3h)=(3h)(3k)

9k+6h=9hk

dividing by 3 we get

3hk-2h-3k=0

Therefore the locus of centroid is

3xy - 2x - 3y =0

Answer 2

Answer:

answer is 3y + 2x = 3xy. ...HOPE IT HELPS YOU. THANK YOU. PLEASE MARK IT AS BRAINLIEST..