KCET · Maths · Straight Lines
\(A B C\) is triangle, \(G\) is the centroid, \(D\) is the mid-point of \(B C\). If \(A=(2,3)\) and \(G=(7,5)\), then the point \(D\) is
- A \(\left(\frac{9}{2}, 4\right)\)
- B \(\left(\frac{19}{2}, 6\right)\)
- C \(\left(\frac{11}{2}, \frac{11}{2}\right)\)
- D \(\left(8, \frac{13}{2}\right)\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{19}{2}, 6\right)\)
Step-by-step Solution
Detailed explanation
Since, \(D\) is the mid-point of \(B C\). So, coordinate of \(B C\) are \(\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)\)
Given, \(G(7,5)\) is the centroid of \(\triangle A B C\).
\(\therefore \quad 7=\frac{2+x_{2}+x_{3}}{3}\) and \(5=\frac{3+y_{2}+y_{3}}{3}\)
\(\Rightarrow \quad x_{2}+x_{3}=21-2\)
and \(\quad y_{2}+y_{3}=15-3\)
\(\begin{array}{ll}\Rightarrow & x_{2}+x_{3}=19 \\ \text { and } & y_{2}+y_{3}=12\end{array}\)
\(\Rightarrow \quad \frac{x_{2}+x_{3}}{2}=\frac{19}{2}\)
and \(\quad \frac{y_{2}+y_{3}}{2}=6\)
\(\therefore\) Coordinate of \(D\) are \(\left(\frac{19}{2}, 6\right)\).
Given, \(G(7,5)\) is the centroid of \(\triangle A B C\).
\(\therefore \quad 7=\frac{2+x_{2}+x_{3}}{3}\) and \(5=\frac{3+y_{2}+y_{3}}{3}\)
\(\Rightarrow \quad x_{2}+x_{3}=21-2\)
and \(\quad y_{2}+y_{3}=15-3\)
\(\begin{array}{ll}\Rightarrow & x_{2}+x_{3}=19 \\ \text { and } & y_{2}+y_{3}=12\end{array}\)
\(\Rightarrow \quad \frac{x_{2}+x_{3}}{2}=\frac{19}{2}\)
and \(\quad \frac{y_{2}+y_{3}}{2}=6\)
\(\therefore\) Coordinate of \(D\) are \(\left(\frac{19}{2}, 6\right)\).
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