Question -2

Geometry Level 4

Given two points A ( 2 , 0 ) , B ( 0 , 4 ) , A(-2,0), B(0,4), find the coordinates of a point M M lying on the line 2 x 3 y = 9 2x-3y=9 so that the perimeter of A M B \triangle AMB is minimized.


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( 21 17 , 37 17 ) \left( \frac{21}{17},\frac{-37}{17} \right) ( 84 13 , 74 13 ) \left( \frac{84}{13},\frac{-74}{13} \right) ( 0 , 3 ) (0,-3) ( 4.5 , 0 ) \left( 4.5, 0\right)

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3 solutions

Rajen Kapur
Feb 28, 2015

First find the mirror reflection of (-2, 0) in the given line 2x - 3y = 9. Perpendicular from (-2, 0) on the line is 3x + 2y + 6 = 0 intersecting at (0, -3). Thus mirror reflection point is (2, -6). Line joining (2, -6) with (0. 4) intersects the given line at (21/17, -37/21) the required point M which makes the perimeter of triangle the least (AB is fixed).
Try my problem "Minimum you can do"

Given pt.doesnot lie on line ?

Anubhav Bhatia - 6 years, 3 months ago

I have updated the answer. Can you update your solution too?

Given the error, please show how the point is determined. Thanks.

Calvin Lin Staff - 6 years, 3 months ago
Guru Prasaadh
Apr 3, 2015

this question is really simple. first they have given a line equation check for all the points in the options u will c that one option automatically gets eliminated. Then think when the perimeter will be the least . if the point m is most nearer to line ab it will be possible. use distance formula between the given line and the options. after reading mine u will see that a level 4 problem doesn't mean that a beginner((exception when he doesn't know the basics properly ))cant do it.just think for some time before u proceed.

And what would you do if this was a subjective question

Spandan Senapati - 4 years, 3 months ago

C h e c k i n g x = 84 13 o n l i n e y = 2 84 13 9 = 3 17 13 3 74 13 . S o t h i s o p t i o n i s r e j e c t e d . D i s t a n c e o f ( 4.5 , 0 ) f r o m A , = 6.5 , f r o m B = 6. T o t a l d i s t a n c e = 12.5. D i s t a n c e o f ( 0 , 3 ) f r o m A , = 13 , f r o m B = 7. T o t a l d i s t a n c e = 10.605555. D i s t a n c e o f ( 21 17 , 37 17 ) f r o m A + f r o m B Checking~x=\dfrac{84}{13}~on ~line~y~=\dfrac{2*84}{13}-9=3\dfrac{17}{13}~\neq~3*\dfrac{-74}{13}.~~So ~this~option ~is ~rejected. \\ Distance~ of~(4.5,0)~from~A,=6.5, ~~from~B=6.~~Total~distance~=12.5.\\ Distance~ of~(0,-3)~from~A,=\sqrt{13}, ~~from~B=7.~~Total~distance~=10.605555.\\ Distance~ of~\left (\dfrac{21}{17},\dfrac{-37}{17} \right )~from~A~+~from ~B\\
= ( 21 17 + 2 ) 2 + ( 37 17 ) 2 + ( 21 17 ) 2 + ( 37 17 4 ) 2 = 10.19. S o t h i s o p t i o n i s c o r r e c t . C h e c k f o r i t t o b e o n 2 x 9 = 3 y . 2 21 17 9 = 2 21 153 17 = 3 37 17 = 3 37 17 . T h i s p o i n t i s t h e s o l u t i o n . =\sqrt{ \left (\dfrac{21}{17}+2 \right )^2+ \left (\dfrac{-37}{17} \right )^2}+\sqrt{ \left (\dfrac{21}{17 } \right )^2+ \left (\dfrac{-37}{17}-4 \right ) ^2}=10.19.\\ \\ So ~this~option~is~correct.~Check ~for~it~to~be~on~2x-9=3y.\\ \dfrac{2*21}{17} -9=\dfrac{2*21-153}{17}=\dfrac{-3*37}{17} =3*\dfrac{-37}{17}.\\ \implies~This~point~is~the~solution.

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