Every point on the curve 3x+2y−3xy=0 is the centroid of a triangle formed by the coordinate axes and a line (L) intersecting both the coordinates axes. Then, all such lines ( L )
Given equation of curve 3x+2y−3xy=0 every point on the curve is the centroid of a triangle formed by the coordinate axes
and a line (L) Since, L intersect the coordinate axes ∴‌ let ‌A‌=(a,0) B‌=(0,b) ∴ Centroid of △OAB, ‌(x,y)=(‌
0+0+a
3
,‌
0+b+0
3
)=(‌
a
3
,‌
b
3
) ‌⇒a=3x,b=3y Equation of line ‌‌
X
3x
+‌
Y
3y
=1 ⇒‌
X
x
+‌
Y
y
=3 Since, centroid (x,y) lies on given curve ‌3x+2y−3xy=0 ‌⇒‌‌3x+2y=3xy ‌⇒‌‌3xy=3x+2y⇒3=‌
3x
xy
+‌
2y
xy
‌⇒‌‌‌
X
x
+‌
Y
y
=‌
3
y
+‌
2
x
‌‌‌ (Using Eq. (i)) ‌ ‌⇒‌‌‌
X−2
x
+‌
Y−3
y
=0 ⇒ Does not represents to a single line Now, 3x+2y=3xy ‌⇒‌‌‌
3x
3xy
+‌
2y
3xy
=1 ‌⇒‌‌‌
1
y
+‌
2
3x
=1 ‌⇒‌‌‌
1
b
3
+‌
2
3(‌
a
3
)
=1 ‌⇒‌‌‌
3
b
+‌
2
a
=1⇒3a+2a=ab ‌⇒ab−3a−2b=0 ‌⇒a=‌
2b
b−3
Consider the equation of line L ‌bX+aY=ab ‌⇒‌‌bX+‌
2b
b−3
Y=b(‌
2b
b−3
) ‌⇒‌‌X+‌
2
b−3
Y=‌
2b
b−3
‌⇒‌‌X(b−3)+2Y=2b ‌⇒‌‌b(X−2)+(−3X+2Y)=0 For all such lines L ‌X−2=0⇒X=2 ‌‌ and ‌‌‌−3X+2Y=0 ‌⇒‌‌Y=3 ⇒‌‌L is passes through (2,3) ⇒‌‌L are concurrent.