Q93 / 160 · AP EAMCET 2015 · Mathematics
- Correct answerA$\begin{array}{llll}\mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} \\ \mathrm{s} & \mathrm{p} & \mathrm{q} & \mathrm{r}\end{array}$
- B$\begin{array}{llll}\mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} \\ \mathrm{p} & \mathrm{q} & \mathrm{r} & \mathrm{s}\end{array}$
- C$\begin{array}{llll}\mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} \\ \mathrm{s} & \mathrm{r} & \mathrm{q} & \mathrm{r}\end{array}$
- D$\begin{array}{llll}\mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} \\ \mathrm{s} & \mathrm{p} & \mathrm{t} & \mathrm{r}\end{array}$
Solution
(A) Given coordinates of triangle are
$\begin{aligned}
& P(2,3,-1), Q(5,6,3), R(2,-3,1) \\
& \therefore \quad \text { Centroid, } G=\left(\begin{array}{c}
\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \\
\frac{z_1+z_2+z_3}{3}
\end{array}\right) \\
& =\left(\frac{2+5+2}{3}, \frac{3+6-3}{3}, \frac{-1+3+1}{3}\right) \\
&
\end{aligned}$
$=\left(\frac{9}{3}, \frac{6}{3}, \frac{3}{3}\right)=(3,2,1)$
(B) Given coordinates of triangle are $P(1,2,3)$, $Q(2,3,1)$ and $R(3,1,2)$
Since, $\triangle P Q R$ is an equilateral triangle.
$\therefore \quad$ Circumcentre of the triangle,
$\begin{aligned}
& C=\left(\begin{array}{r}
\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \\
\frac{z_1+z_2+z_3}{3}
\end{array}\right) \\
& =\left(\frac{1+2+3}{3}, \frac{2+3+1}{3}, \frac{3+1+2}{3}\right)=(2,2,2)
\end{aligned}$
(C) Given coordinates of triangle are $P(2,1,5)$, $Q(3,2,3)$ and $R(4,0,4)$.
Since, $\triangle P Q R$ is an equilateral triangle.
$\therefore \quad$ Orthocentre of the triangle,
$\begin{aligned}
& O=\left(\begin{array}{c}
\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \\
\frac{z_1+z_2+z_3}{3}
\end{array}\right) \\
& =\left(\frac{2+3+4}{3}, \frac{1+2+0}{3}, \frac{5+3+4}{3}\right)=(3,1,4)
\end{aligned}$
(D) Given coordinates of triangle are $P(0,0,0)$, $Q(3,0,0)$ and $R(0,4,0)$.
Since, $\triangle P Q R$ is a right angled triangle.
Here, $a=5, b=4, c=3$
$\therefore \quad$ Incentre of the triangle,
$I=\left(\begin{array}{c}
\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c}, \\
\frac{a z_1+b z_2+c z_3}{a+b+c}
\end{array}\right)$
$\begin{aligned}
& =\left(\begin{array}{l}
\frac{5 \times 0+4 \times 3+3 \times 0}{3+4+5} \\
\frac{5 \times 0+4 \times 0+3 \times 4}{3+4+5} \\
\frac{5 \times 0+4 \times 0+0 \times 3}{3+4+5}
\end{array}\right) \\
& =\left(\frac{12}{12}, \frac{12}{12}, 0\right)=(1,1,0)
\end{aligned}$
Hence, the correct answer is option: 'A'.