TS EAMCET · Maths · Straight Lines
If \(A(4,3,5), B(0,-2,2)\) and \(C(3,2,1)\) are three points. Then, the coordinates of the point in which the bisector of \(\angle B A C\) meets the side \(B C\), is
- A \(\left(\frac{15}{8}, \frac{4}{8}, \frac{11}{8}\right)\)
- B \(\left(\frac{12}{7}, \frac{2}{7}, \frac{10}{7}\right)\)
- C \(\left(\frac{9}{5}, \frac{2}{5}, \frac{7}{5}\right)\)
- D \(\left(\frac{3}{2}, 0, \frac{3}{2}\right)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{15}{8}, \frac{4}{8}, \frac{11}{8}\right)\)
Step-by-step Solution
Detailed explanation
We know that angle bisector of \(\angle B A C\) meets \(B C\) at \(D\) which divides \(B C\) in the ratio \(A B: A C\). i.e. \(\quad \frac{B D}{C D}=\frac{A B}{A C}\)…
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