The centroid of the triangle formed by the pair of lines 12x^2-20xy+7y^2=0 and the line 2x-3y+4=0 is

1)[7/3,-7/3]
2)[8/3,8/3]
3)[-4/3,-4/3]
4)none of these

the given pair is
12x^2-20xy+7y^2=0
(6x-7y)(2x-y)=0
y=6/7x,y=2x
so,the lines forming the triangle are
y=2x
y=6/7x
and 2x-3y+4=0
0n solving above equations we find the verticesof triangle as O(0,0),A(7,6),B(1,2)
Hence,the centroid of the triangle is
[(0+7+1)/3,(0+6+2)/3]= [8/3,8/3]

12x² - 20xy + 7y² = 0
(2x - y)(6x - 7y) = 0

The first equation represents these two lines:

2x - y = 0
6x - 7y = 0

The equations intersect at (0, 0). Now see where the other equation intersects each of them.

2x - y = 0
y = 2x
2x - 3y + 4 = 0
y - 3y + 4 = 0
y = 2
x = 1
(1, 2)

2x - 3y + 4 = 0
6x - 9y + 12 = 0
6x - 7y = 0
2y - 12 = 0
y = 6
x = 7
(7, 6)

The three points of intersection are (0, 0), (1, 2), and (7, 6).
mean of x-coordinates = 8/3
mean of y-coordinates = 8/3
Centroid: (8/3, 8/3)