The coordinates of the centroid of the triangle, whose sides are 12x2-20xy+7y2=0 and 2x-3y+4=0 is 83,83. - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

12 x^{2} - 20 xy + 7 y^{2} = 0

\Rightarrow (6 x - 7 y) (2 x - y) = 0

\Rightarrow y = \frac{6 x}{7}, 2 x

The intersection points of the line 2x−3y+4=0 with pair of straight lines

12 x^{2} - 20 xy + 7 y^{2} = 0

are
(1, 2) and (7, 6).
The vertices of the triangle as A(0,0), B(1,2) and C(7,6).
Centroid of triangle ABC is

(\frac{0 + 1 + 7}{3}, \frac{0 + 2 + 6}{3}) = (\frac{8}{3}, \frac{8}{3})

.
Hence, the given statement is true.

The coordinates of the centroid of the triangle, whose sides are $12 x^{2} - 20 xy + 7 y^{2} = 0$ and $2 x - 3 y + 4 = 0$ is $(\frac{8}{3}, \frac{8}{3})$ .