Question

The vertices $P,Q$ and $R$ of a triangle are $(2,1),(5,2)$ and $(3,4)$, respectively. Then, the circumcenter is

• A. $\left(\frac{13}{4},-\frac{9}{4}\right)$
• B. $\left(-\frac{13}{4},\frac{9}{4}\right)$
• C. $\left(-\frac{13}{4},-\frac{9}{4}\right)$
• D. $\left(\frac{13}{4},\frac{9}{4}\right)$

Answer 1

The correct option is D

$\left(\frac{13}{4},\frac{9}{4}\right)$

Explanation of the correct option:

Step-1 Equation of perpendicular :

Given: $P,Q$ and $R$ are the vertices of the triangle.

Let $A(x,y)$ be the circumcenter of triangle $△PQR$.

Since, $AP,AQ,AR$ are the radii of the circumcenter.

Therefore, $AP=AQ=AR$

Taking, $AP=AQ$

$A{P}^2=A{Q}^2$

$\Rightarrow$${\left(2-x\right)}^2+{(1-y)}^2={(5-x)}^2+{(z-y)}^2$

$\Rightarrow$ $6x+2y=24$…………….$\left(1\right)$

Similarly, $A{P}^2=A{R}^2$

$\Rightarrow$${\left(2-x\right)}^2+{(1-y)}^2={(3-x)}^2+{(4-y)}^2$

$\Rightarrow$ $x+3y=10$

$\Rightarrow$ $x=10-3y$………$\left(2\right)$

Step-2 : Circumcentre of triangle :

Substitute equation $\left(2\right)$ in equation $\left(1\right)$,

$$\begin{gathered} 6\left(10-3y\right)+2y=24 \\ \Rightarrow 60-18y+2y=24 \\ \Rightarrow 60-24=16y \\ \Rightarrow y=\frac{36}{16} \\ \Rightarrow y=\frac{9}{4} \\ \Rightarrow x+\frac{27}{4}=10 \\ \Rightarrow x=\frac{13}{4} \end{gathered}$$

Putting the value of $x$ in equation $\left(2\right)$,

$$\begin{gathered} \frac{13}{4}=10-3y \\ \Rightarrow 3y=10-\frac{13}{4} \\ \Rightarrow 3y=\frac{40-13}{4} \\ \Rightarrow 3y=\frac{27}{4} \\ \Rightarrow y=\frac{27}{3\times 4} \\ \Rightarrow y=\frac{9}{4} \end{gathered}$$

Therefore, the circumcenter of a triangle $PQR$ is $\left(\frac{13}{4},\frac{9}{4}\right)$.

Hence option D is the correct option.