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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: (fast please) locus of a point........
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sinjan.j (574)

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two points A and B with coordinates (5,3)and(3,-2) are given.
A point P moves so that the area of
PAB is constant and equal to 9 sq.units.

find the equation of locus of P.

guys i don't want u to solve it for me. but i only want to know about the formula (related to the triangle) ..... so that i can solve it myself.

please make it fast.....




    
coolank2 (126)

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I think the ans is
5x - 11y = 28

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sinjan.j (574)

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coolank the answer is wrong.....

the correct answer is 5x-2y-37=0





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sinjan.j (574)

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by the way i asked u to give me the procedure pal




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srujana (3045)

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Use the formula Area of a triangle = 1/2 |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|
 
x1,y1= (5,3)
 
x2,y2=(3,-2)
 
x3,y=(x,y)
 
 

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coolank2 (126)

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Sorry yaar ......
Made a calculation mistake ....
 
Solve it by determinant method ....
There can be 2 answers actually .....
 
5x - 2y = 1 and 5x - 2y = 37
 
Sorry once again ......

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sinjan.j (574)

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thanx guys for the help...............!!!!




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sinjan.j (574)

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1 more question...

coolank

how do u get 2 answers...???




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srujana (3045)

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Sinjan, modulus sign gives 2 ans because |x| may mean + or - x

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make yourself another.
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coolank2 (126)

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Ya, Srujana is right .......

This is just da beginning .....
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sinjan.j (574)

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ok fine thanx srujana




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sinjan.j (574)

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hey can someone work it out for me i am not getting the answer...!!!




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coolank2 (126)

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It would be

(1/2) mod ( | 5 3 1 | ) = 9
| 3 -2 1 |
| x y 1 |

You'll get it .........
Could not get it more legible ........

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srujana (3045)

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1/2 |5(-2-y)+3(y-3)+x(3+2)|  = 9
 
|-10-5y+3y-9+5x|
 
5x-2y-19= + or -18
 
5x-2y=37
 
or
 
5x-2y=1

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make yourself another.
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You were born an original. Don't die a copy.
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