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BITSAT 2024Paper 1
Exam
BITSAT
Year
2024
Paper
Paper 1
Subject
Mathematics
Chapter
Pair of Lines

BITSAT 2024 · Paper 1 · Mathematics · Pair of Lines

A(3,2,0), B(5,3,2), C(-9,6,-3) are three points forming a triangle....

A(3,2,0),B(5,3,2),C(−9,6,−3)\mathrm{A}(3,2,0), \mathrm{B}(5,3,2), \mathrm{C}(-9,6,-3) are three points forming a triangle. AD , the bisector of angle BAC meets BC in D. Find the co-ordinates of D.A(3,2,0), B(5,3,2), C(-9,6,-3) are three points forming a triangle. AD , the bisector of angle BAC meets BC in D. Find

Options

  1. A198,5715,5715\frac{19}{8}, \frac{57}{15}, \frac{57}{15}
  2. B198,5716,1716\frac{19}{8}, \frac{57}{16}, \frac{17}{16}
  3. C(2,3,0)(2,3,0)
  4. D(4,5,6)(4,5,6)

Correct answer

B. 198,5716,1716\frac{19}{8}, \frac{57}{16}, \frac{17}{16}

Step-by-step solution

Since ADA D is the bisector of ∠BAC\angle B A C. ⇒BDDC=ABAC\Rightarrow \frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}} ....(i) Now, AB=(5−3)2+(3−2)2+(2−0)2\mathrm{AB}=\sqrt{(5-3)^2+(3-2)^2+(2-0)^2} =4+1+4=9=3=\sqrt{4+1+4}=\sqrt{9}=3 A(3,2,0), B(5,3,2), C(-9,6,-3) are three points forming a triangle. AD , the bisector of angle BAC meets BC in D. Find AC=(−9−3)2+(6−2)2+(−3−0)2A C=\sqrt{(-9-3)^2+(6-2)^2+(-3-0)^2} =144+16+9=13=\sqrt{144+16+9}=13 ∴BDDC=313\therefore \quad \frac{\mathrm{BD}}{\mathrm{DC}}=\frac{3}{13} Hence, D divides BC in the ratio 3:133: 13 The co-ordinates of D are (3(−9)+13(5)3+13,3(6)+13(3)3+13,3(−3)+13(2)3+13)=(198,5716,1716)\left(\frac{3(-9)+13(5)}{3+13}, \frac{3(6)+13(3)}{3+13}, \frac{3(-3)+13(2)}{3+13}\right)=\left(\frac{19}{8}, \frac{57}{16}, \frac{17}{16}\right)
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