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Most Important Selected Qs for JEE AdvancedMathematicsStraight Lines

Paragraph: Straight lines 3 x+4 y=5 and 4 x-3 y=15 intersect at A . Points B and C are chosen on these lines such that AB = AC .(1,2) is a point on the line BC . Distance of BC from A is less than 2 . On the basis of above information answer the following : Question: The area of ABC is -

Options

  1. A2.24
  2. B2.42
  3. C2
  4. D11 5 2

Correct answer

B. 2.42

Step-by-step solution

Line BC will be parallel to angle bisectors of angle A (3,-1) so 3 x+4 y-5 5 = 4 x-3 y-15 5 by +ve sign x-7 y-10=0 . B₁ -ve sign 7 x + y -20=0 . B ₂ So B C can be x-7 y= ₁ ₁=-13 or 7 x + y = ₂ ₂=9 so x-7 y=-13 or 7 x+y=9 Distance of x-7 y=-13 from A(3,-1) is 23 50 2 and distance of 7 x+y=9 from A (3,-1) is 11 5 2 1 so BC is 7 x + y =9 ABC is right isosceles triangle so area of A B C= 1 2 2 = ^2= ( 11 5 2 )^2= 121 50

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