Question

If $A(x_1,y_1),B{x}_2,y_2),C(x_3,y_3)$ are the vertices of an equilateral triangle and $(x_1-2{)}^2+(y_1-3{)}^2=(x_2-2{)}^2+(y_2-3{)}^2=(x_3-2{)}^2+(y_3-3{)}^2$, then

• A. Coordinates of orthocentre is $(0,0)$
• B. Coordinates of circumcentre is $(2,3)$
• C. value of $(x_1+x_2+x_3)+3(y_1+y_2+y_3)$ is $33$
• D. value of $2(x_1+x_2+x_3)-(y_1+y_2+y_3)$ is $3$

Answer 1

Answer: (B), (C), (D)

The correct options are
B Coordinates of circumcentre is $(2,3)$
C value of $(x_1+x_2+x_3)+3(y_1+y_2+y_3)$ is $33$
D value of $2(x_1+x_2+x_3)-(y_1+y_2+y_3)$ is $3$

Here, $△ABC$ is equilateral
$(x_1-2{)}^2+(y_1-3{)}^2=(x_2-2{)}^2+(y_2-3{)}^2=(x_3-2{)}^2+(y_3-3{)}^2$
$\Rightarrow A,B,C$ are equidistant from the point $(2,3)$
$\therefore (2,3)$ is the circumcentre
Also $(2,3)$ will be the coordinates of centroid and orthocentre.
$(\because$ in equilateral triangle circumcentre, orthocentre and centroid coincides with each other $)$

Now
$(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})=(2,3)\Rightarrow x_1+x_2+x_3=6,y_1+y_2+y_3=9(x_1+x_2+x_3)+3(y_1+y_2+y_3)=332(x_1+x_2+x_3)-(y_1+y_2+y_3)=3$