MathematicsThe points A(2,9),B(a,5)   and C(5,5)   are the vertices of ΔABC   right angled at B  . Find the value of a   and hence the area of ΔABC   from the following choices is:

The points A(2,9),B(a,5)   and C(5,5)   are the vertices of ΔABC   right angled at B  . Find the value of a   and hence the area of ΔABC   from the following choices is:


  1. A
    a=(0,5)   and Area =6   square units
  2. B
    a=(2,0)   and Area =6   square units
  3. C
    a=(2,5)   and Area =6   square units
  4. D
    a=(2,5)   and Area =6   square units 

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    Solution:

    Given that, the points  A(2,9),B(a,5)   and C(5,5)  .
    The distance formula between two points ( x 1 , y 1 ),( x 2 , y 2 )   is given by,
    D= x 2 x 1 2 + y 2 y 1 2  
    By using Pythagoras theorem:
    B C 2 +A B 2 =A C 2  
    Distance between the points AB:
    Here,
    ( x 1 , y 1 )=(a,5) ( x 2 , y 2 )=(5,5)  
    Now, BC= a5 2 + 55 2 BC= a5 2   Distance between the points AB:
    Here,
    ( x 1 , y 1 )=(2,9) ( x 2 , y 2 )=(a,5)  
    AB= a2 2 + 59 2 AB= a2 2 +16  
    Distance between the points AC:
    Here,
    ( x 1 , y 1 )=(2,9) ( x 2 , y 2 )=(a,5)  
    AC= 52 2 + 59 2 AC=25  
    Substitute the value in the Pythagoras theorem,
    B C 2 +A B 2 =A C 2 a5 2 + a2 2 +16=25 a 2 4a+4+ a 2 10a+25=9 2 a 2 14a+20=0   By simplification,
    a 2 7a+10=0 a 2 5a2a+10=0 a a5 2 a5 =0 a2 a5 =0   a=2&a=5 Area= 1 2 ×b×h Area= 1 2 ×3×4 Area=6 square units  
    The value of a=2&a=5  (2, 5) and the area is 6   square units.
    Hence, option 4) is correct.
     
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