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The points A(2,9),B(a,5)and c(5,5)are the vertices of a triangle ABC,right angled at B.Find the value of a and hence the area of triangle ABC.
The points A(2, 9), B(a, 5) C(5, 5) are the vertices of a triangle ABC. Since it a right triangle, right angled at B
therefore , AB2 + BC2 = AC2
AB2 = (a - 2)2 + (5 - 9)2 = (a - 2)2 + 16
BC2 = (5 - a)2 + (5 - 5)2 = (5 - a)2
AC2 = ( 5 - 2 )2 + ( 5 - 9 )2 = 9 + 19 = 25

ON PUTTING THE VALUES,
AC2 = AB2 + BC2
25 = ( a - 2 )2 + 16 + ( 5 - a )2
25 = a2 + 4 - 4a + 16 + 25 + a2 - 10a
25 = 2a2 - 14a + 45
0 = 2a2 - 14a + 45 - 25
0 = 2a2 - 14a + 20
0 = 2 (a2 - 7a +10)
0 = a2 - 5a -2a +10
0 = a (a - 5) - 2 (a - 5)
0 = ( a - 5 ) ( a - 2 )
a = 5 or 2 ( the value of a cannot be 5 as in that case the coordinates of points B and C will be same and they will coincide. )
Hence, a = 2
AB2 = ( 2 - 2 )2 + 16 = 16
AB = 4
BC2 = ( 5 - A )2
BC = 5 - A
BC = 5 - 2
BC = 3
area of the triangle ABC = 1/2 *base * height
= 1/2 x 3 x 4
= 1/2 x 12
= 6 sq. units
a = 2, 5 and area = 6 sq. units


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