Question

Orthocentre of a triangle is $(2,1)$ and the circumcentre is $(\frac{7}{2},\frac{5}{2})$ then its nine points circle centre is..

• A. $(\frac{11}{4},\frac{7}{4})$
• B. $(\frac{11}{2},\frac{7}{4})$
• C. $(\frac{7}{2},\frac{11}{2})$
• D. $(\frac{7}{4},\frac{11}{4})$

Answer 1

Answer: (A)

$(\frac{11}{4},\frac{7}{4})$

$\begin{array}{cc}\text{Solution -}& \text{Orthocenter}=(2,1)\\ & \text{Circumcenter}=(\frac{7}{2},\frac{5}{2})\end{array}$ $\begin{array}{cc}\Rightarrow x& =\frac{2+\frac{7}{2}}{2}=\frac{11}{4}\\ y& =\frac{1+5/2}{2}=\frac{7}{4}\end{array}$

: nine-point circle center $=(x,y)=(\frac{11}{4},\frac{7}{4})$

Hence (A) is the correct option.