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: Circumcentre of ABC is -
Options
- A( 73 25 ,- 61 25 )
- B( 73 14 ,- 31 14 )
- C( 24 7 ,- 3 7 )
- Dnone of these
Correct answer
A. ( 73 25 ,- 61 25 )
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 Now for 'B' 3 x+4 y=57 x+y=9-25 x=-31 x= 31 25 y= 8 25 B ( 31 25 , 8 25 ) Now for 'C' 4 x-3 y=157 x+y=925 x=42 x= 42 25 y=- 69 25 So C ( 42 25 ,- 69 25 ) Circumcentre will be mid poin