UPSC NDA I 2015 Mathematics Solved Paper

The required co-ordinates of the triangle are as follows:
(3,4), (5,2), (a, 5)
Therefore we can find the area bounded by the triangle using determinant method.
Hence
=

1

2

[(25−15−14)−2a+4a]
=

1

2

[−4+2a]
Now it is given that the area of the triangle so formed is 3 sq. units.
Hence
3=

1

2

|2a−4|
|2a−4|=6
2a−4=±6
2a=10,2a=−2
a=5,a=−1
Therefore,one of the values of 'a' is 5