Question

(xi, yi) are vertices of a equilateral triangle ABC such that $(x_1-2{)}^2+(y_1-3{)}^2=(x_2-2{)}^2+(y_2-3{)}^2=(x_3-2{)}^2+(y_3-3{)}^2.then2(x_1+x_2+x_3(y_1+y_2+y_3)$ =

• A. 30
• B. 29
• C. 39
• D. None of theae

Answer 1

Answer: (D)

None of theae

$(x_i,y_i)$ are vertical of a equilateral triangle.

$(x_1-2{)}^2+(y_1-3{)}^2=(x_2-2{)}^2+(y_2-3{)}^2=(x_3-2{)}^2+(y_3-3{)}^2$

Let, $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ be the co-ordinates of triangle ABC.

Let, P be a point whose coordinate is ( 2 , 3 )

Hence, distance between two points is given by $=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$

$\therefore$ Distance between A & P $=\sqrt{(x_1-2{)}^2+(y_1-3{)}^2}$

$\therefore$ Distance between B & P $=\sqrt{(x_2-2{)}^2+(y_2-3{)}^2}$

$\therefore$ Distance between C & P $=\sqrt{(x_3-2{)}^2+(y_3-3{)}^2}$

Now since , ABC in equilateral triangle, P is centroid or orthocenter of it.

Now, we know centroid co-ordinates is given by $(\frac{x_1,+x_2+y_3}{3},\frac{y_1+y_2+y_3}{3})$

Hence P (2,3) & P $(\frac{x_1,+x_2+y_3}{3},\frac{y_1+y_2+y_3}{3})$

Comparing both we get,

$2=\frac{x_1,+x_2+y_3}{3},3=\frac{y_1+y_2+y_3}{3}$

$\Rightarrow x_1,+x_2+y_3=6,y_1+y_2+y_3=9$

Hence, $2(x_1,+x_2+y_3)+(y_1+y_2+y_3)=(2\times 6)+9=21$

$\therefore$ option D is correct.