KCET · Maths · Three Dimensional Geometry
The mid points of the sides of triangle are \((1,5,-1)(0,4,-2)\) and \((2,3,4)\) then centroid of the triangle
- A \((1,4,3)\)
- B \(\left(1,4, \frac{1}{3}\right)\)
- C \((-1,4,3)\)
- D \(\left(\frac{1}{3}, 2,4\right)\)
Answer & Solution
Correct Answer
(B) \(\left(1,4, \frac{1}{3}\right)\)
Step-by-step Solution
Detailed explanation
Given, mid points of the sides of triangle are \((1,5,-1),(0,4,-2)\) and \((2,3,4)\).
Let, \(\left(x_{1}, y_{1}, z_{1}\right)=(1,5,=1)\),
\(\left(x_{2}, y_{2}, z_{2}\right)=(0,4,-2) \text {, }\)
\(\left(x_{3}, y_{3}, z_{3}\right)=(2,3,4) \text {. }\)
Then centroid of triangle, \(G(x, y, z)=\)
\(\begin{aligned}
&\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3}\right) \\
&G(x, y, z)=\left(\frac{1+0+2}{3}, \frac{5+4+3}{3}, \frac{-1-2+4}{3}\right) \\
&=\left(1,4, \frac{1}{3}\right)
\end{aligned}\)
Let, \(\left(x_{1}, y_{1}, z_{1}\right)=(1,5,=1)\),
\(\left(x_{2}, y_{2}, z_{2}\right)=(0,4,-2) \text {, }\)
\(\left(x_{3}, y_{3}, z_{3}\right)=(2,3,4) \text {. }\)
Then centroid of triangle, \(G(x, y, z)=\)
\(\begin{aligned}
&\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3}\right) \\
&G(x, y, z)=\left(\frac{1+0+2}{3}, \frac{5+4+3}{3}, \frac{-1-2+4}{3}\right) \\
&=\left(1,4, \frac{1}{3}\right)
\end{aligned}\)
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