Question

Match the following
I. The centroid of the triangle formed by (2, 3, -1), (5,6,3), (2, - 3, 1) is
II The circumcentre of the triangle formed by (1,2,3), (2,3,1), (3,1,2) is
III The orthocentre of the triangle formed by (2,1,5), (3,2,3), (4,0,4) is
IV The incentre of the triangle formed by (0,0,0), (3,0,0), (0,4,0) is
A) (2,2,2)
B) (3,1,4)
C) (1,1,0)
D) (3,2,1)
E) (0,0,0)

(a) I-D, II-A, III-B, IV-C
(b) I-A, II-B, III-C, IV-D
(c) I-D, II-E, III-B, IV-C
(d) I-D, II-A, III-E, IV-C

Answer 1

Hint: For solving this problem, we will individually consider each case and try to find the desired coordinates of the triangle as mentioned in the problem. After finding each coordinate, we can easily match with the given option to obtain the final answer.

Complete step-by-step answer:
Considering part (I), the formula for finding the coordinate of centroid of triangle is $\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3},\dfrac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3} \right)$.
Now, the centroid of the triangle formed by (2, 3, -1), (5,6,3), (2, - 3, 1):
$\begin{aligned} & \left( \dfrac{2+5+2}{3},\dfrac{3+6-3}{3},\dfrac{-1+3+1}{3} \right) \\\ & \Rightarrow (3,2,1) \\\ \end{aligned}$
By analysing the options, option (b) is eliminated.

Considering part (II), the triangle formed by (1,2,3), (2,3,1), (3,1,2) is an equilateral triangle because by using the distance formula, all the side length is equal to $\sqrt{6}$.
We know that distance between two points is given by ,
$D$=${\sqrt{{{(x_2-x_1)}^{2}}+{{(y_2-y_1)}^{2}}+{{(z_2-z_1)}^{2}}}}$

& \because \sqrt{{{(2-1)}^{2}}+{{(3-2)}^{2}}+{{(1-3)}^{2}}}=\sqrt{6} \\\ & \sqrt{{{(3-2)}^{2}}+{{(1-3)}^{2}}+{{(2-1)}^{2}}}=\sqrt{6} \\\ & \sqrt{{{(1-3)}^{2}}+{{(2-1)}^{2}}+{{(3-2)}^{2}}}=\sqrt{6} \\\ \end{aligned}$$ So, for an equilateral triangle the circumcentre and centroid are the same. By using the formula of centroid, we get $\begin{aligned} & \left( \dfrac{1+2+3}{3},\dfrac{2+3+1}{3},\dfrac{3+1+2}{3} \right) \\\ & \Rightarrow (2,2,2) \\\ \end{aligned}$ By analysing the options, option (c) is eliminated. So, now we are left with only two options. Considering part (III), let the orthocentre of the triangle A (2,1,5), B (3,2,3), C (4,0,4) be P (x, y, z). Now, by the definition of orthocentre, $AP\bot BC,BP\bot AC,CP\bot AB$. The direction ratios are defined as the difference in the coordinates of two points i.e., $({{x}_{2}}-{{x}_{1}},{{y}_{2}}-{{y}_{1}},{{z}_{2}}-{{z}_{1}})$ So, direction ratios of AP = (x-2, y-1, z-5). The direction ratios of BC = (4-3,0-2,4-3) = (1, -2,1). To evaluate this part in a quick manner, we use the fact that the product of direction ratios of two perpendicular lines is zero. So, by using the property of perpendicular direction ratios: $\begin{aligned} & \left( x-2 \right)\cdot 1+(y-1)\cdot (-2)+(z-5)\cdot 1=0 \\\ & x-2-2y+2+z-5=0 \\\ & x-2y+z=5 \\\ \end{aligned}$ Therefore, orthocentre lies in this equation. Now, satisfying all the coordinates as mentioned in question, we obtain that only (3,1,4) satisfies the equation. Hence, orthocentre is (3,1,4). Hence, option (d) is also eliminated. Therefore, the correct option is option (a). Note: This problem can alternatively be solved for all the four parts and then option for the corresponding answer could be marked. But on solving all the four parts, more time is required. So, to cut short the time, the above strategy is useful.