Question

The distance between the circumcentre and the centroid of the triangle formed by the vertices $(1,2),(3,−1)$ and $(4,0)$ is

• A. $\frac{1}{\sqrt{2}}\sqrt{45}$
• B. $4$
• C. $\frac{7\sqrt{2}}{15}$
• D. $\frac{11\sqrt{2}}{30}$

Answer 1

Answer: (D)

We have,
$A(1,2),B(3,−1),C(4,0)$
Centroid of $ΔABC$ is

$\left(\frac{1+3+4}{3},\frac{2−1+0}{3}\right)=\left(\frac{8}{3},\frac{1}{3}\right)$
Let $(h,k)$ be circumcentre of $ΔABC$
$∴OA=OB=OC,OA=OB$
$(h−1{)}^2+(k−2{)}^2=(h−3{)}^2+(k+1{)}^2$
$⇒h^2−2h+1+k^2−4k+4=h^2−6h+9+k^2+2k+1$
$⇒4h−6k=5$ ...(i)
Similarly, $OA=OC$
$âˆ\mu (h−1{)}^2+(k−2{)}^2=(h−4{)}^2+k^2$
$⇒6h−4k=11$ ...(ii)
Solving Eqs. (i) and (ii), we get
$h=\frac{23}{10},k=\frac{7}{10}$
Distance between circumcentre and centroid of $â–³ABC$ is
$\sqrt{{\left(\frac{23}{10}−\frac{8}{3}\right)}^2+{\left(\frac{7}{10}−\frac{1}{3}\right)}^2}=\frac{11\sqrt{2}}{30}$