Question

Centroid of a triangle, the equation of whose sides are $12{x^2} - 20xy + 7{y^2} = 0$ and $2x - 3y + 4 = 0$ is
$A.{\text{ }}\left( {\dfrac{8}{3},\dfrac{8}{3}} \right) \\\ B.{\text{ }}\left( {\dfrac{4}{3},\dfrac{4}{3}} \right) \\\ C.{\text{ }}\left( {2,2} \right) \\\ D.{\text{ }}\left( {1,1} \right) \\\$

Answer 1

Hint- In this question, firstly dissociate the pair of straight lines to two different equations of straight line and find the coordinates of point of intersection of three straight lines taking two at a time. This will give coordinates of vertices of the triangle to find the centroid.

Complete step-by-step answer:

Equation of pair of straight line
$12{x^2} - 20xy + 7{y^2} = 0$
Break the middle term or $xy$ term to convert it into two multiples
$12{x^2} - 6xy - 14xy + 7{y^2} = 0$
$6x(2x - y) - 7y(2x - y) = 0 \\\ (6x - 7y)(2x - y) = 0 \\\$
$\therefore$Sides of triangle are
$6x - 7y = 0...............(1) \\\ 2x - y = 0..................(2) \\\ 2x - 3y + 4 = 0...........(3) \\\$
From equation (1) and (2)

$$6x - 7y = 0...............(1) \\\ 2x - y = 0..................(2) \\\$$

Multiply (2) by 3 and subtract it from (1),we get
$\- 7y + 3y = 0 \\\ \- 4y = 0 \\\ y = 0 \\\$
Put this value of y in (1), we get
$x = 0$
So,$x = 0,y = 0$
From equation (2) and (3)

$$2x - y = 0..................(2) \\\ 2x - 3y + 4 = 0...........(3) \\\$$

Subtract (1) from (2), we get
$\- 2y + 4 = 0 \\\ \- 2y = - 4 \\\ y = \dfrac{4}{2} = 2 \\\$
Put this value of y in (2), we get
$2x - 2 = 0 \\\ 2x = 2 \\\ x = 1 \\\$
So,$x = 1,y = 2$
From equation (1) and (3)
$6x - 7y = 0...............(1) \\\ 2x - 3y + 4 = 0...........(3) \\\$
Multiply (3) with 3 and subtract it from (1), we get
$2y - 12 = 0 \\\ 2y = 12 \\\ y = \dfrac{{12}}{2} = 6 \\\$
Put this value of y in (1), we get
$6x - 7 \times 6 = 0 \\\ x = \dfrac{{7 \times 6}}{6} = 7 \\\$
So,$x = 7,y = 6$

Hence coordinates of centroid is
$\left( {\dfrac{{1 + 7 + 0}}{3},\dfrac{{2 + 6 + 0}}{3}} \right) \\\ = \left( {\dfrac{8}{3},\dfrac{8}{3}} \right) \\\$
Therefore, the correct option is A.

Note- Centroid of a triangle is the point of intersection of three medians of the triangle (connecting a vertex and midpoint of opposite side). If $({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})$are coordinates of three vertices of triangle then coordinates of centroid is $\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$.