$\begin {array} {1 1} (A)\;(-1,-1) & \quad (B)\;(2,2) \\ (C)\;(-2, -2) & \quad (D)\;(2,-2) \end {array}$

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- If two lines are perpendicular then the product of their slopes is $m_1m_2=-1$
- The centroid of a triangle ABC divides the line AD in the ratio 2 : 1.

Step 1 :

Let ABC be the triangle and let the coordinates of A be (h, k).

Let D be any point on BC and let the coordinates of D be ( $ \alpha, \beta $).

It is given that the coordinates of the centroid $O$ is (0,0)

$Ao : oD $ is 2 : 1

$ \therefore \large\frac{2\alpha+h}{2+1}$$=0 $ and $ \large\frac{2 \beta + k}{2+1}$$=0$

$ \Rightarrow 2 \alpha + h =0$ and $ 2\beta + k = 0$

$ \therefore 2 \alpha + h = 2 \beta + k$

$ \Rightarrow 2( \alpha - \beta ) = k - h$---------(1)

Also given BC = $x+y-2=0$ Since it passes through $( \alpha, \beta ); \alpha + \beta -2 = 0$-------(2)

Slope of the line AD is $ \large\frac{k-0}{h-0}$$ = \large\frac{k}{h}$

Slope of the line BC $x+y-2$ is -1

$ \therefore \large\frac{k}{h}$$ \times -1 = -1$

$ \Rightarrow k = h $-------(3)

$ \therefore \alpha - \beta = 0$

$ \alpha + \beta = 2$

Solving the above equations we get,

$ 2\alpha = 2$

$ \alpha = 1 $ and $ \beta = 1$

Substituting for $ \alpha $ and $ \beta $ we get,

$ 2 \alpha + h = 0$

$ 2(1) +h=0 \qquad \Rightarrow h = -2$

and $2\beta +k =0$

$ 2(1) + k = 0 \qquad \Rightarrow k = -2$

Hence the coordinates of the vertex A is (-2,-2).

Hence the correct option is 'C'.

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