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1. a variable straight line passes through a fixed point (a,b) intersecting the co-ordinate axies at A & B. if O is the origin then the locus of the centroid of the triangle OAB is- a. bx +ay -3xy=0 b. bx +ay -2xy=0 c. ax +by -3xy=0 d. ax +by -2xy=0

1. a variable straight line passes through a fixed point (a,b) intersecting the co-ordinate axies at A & B. if O is the origin then the locus of the centroid of the triangle OAB is-


a. bx +ay -3xy=0


b. bx +ay -2xy=0


c. ax +by -3xy=0


d. ax +by -2xy=0

Grade:11

2 Answers

vikas askiitian expert
509 Points
10 years ago

the equation of line is

(y-b) = m(x-a)

here m is variable & line passes through (a,b)..

intercept on y axis = (0,b-am)

intercept on x axis = (a -b/m , 0)

onigin = (0,0)

centroid = (x,y) = [ (0+a - b/m + 0)/3  ,  (b-am + 0+0)/3]

3x = a - b/m          ..............1

3y = b - am           ..............2

after solving these equatin we get

bx + ay - 3xy = 0

option a) is correct

approve my ans if u like

Mohini
26 Points
3 years ago
As the line intersects the axes at A & B ,these will be our y & x intercepts ,so let the points be : A(0,k) & B(h,0) 
 
so equation of the line : x/h + y/k =1  (intercept form)
 
Also it passes through (a,b), therefore, putting it in equation, we have: a/h +b/k =1 …..........1
 
Now, centroid=(x,y)=[(0+0+h)/3 , (0+k+0)/3]  ,this gives: h=3x & k=3y  ,, now putting the value of h & k in above equation 1
 
a/3x + b/3y =1 ,,which gives   bx + ay = 3xy

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