Find the number of integral points in the interior of the triangle having vertices and .
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We observe that there are a total of 2 ( n − 2 ) ( n − 1 ) = 6 integral points, let us call this equation 1
now we know the pattern and the foramula required then it can easily be done for ( 0 , 0 ) , ( 0 , 2 1 ) a n d ( 2 1 , 0 )
substituting n = 2 1 in 1
we get: 2 ( 1 9 ) ( 2 0 ) = 1 9 0