The number of points,having both coordinates | vedclass

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(A) The vertices of the triangle are $(0,0)$,$(41,0)$,and $(0,41)$.The equation of the line passing through $(41,0)$ and $(0,41)$ is $x + y = 41$.We n

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$780$

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(A) The vertices of the triangle are $(0,0)$,$(41,0)$,and $(0,41)$.The equation of the line passing through $(41,0)$ and $(0,41)$ is $x + y = 41$.We need to find the number of integer pairs $(x, y)$ such that $x > 0$,$y > 0$,and $x + y For a fixed $x$,the possible values for $y$ are $1, 2, \dots, 40-x$.The number of such points for a given $x$ is $40-x$.Summing this from $x = 1$ to $39$:$\sum_{x=1}^{39} (40-x) = 39 + 38 + \dots + 1$.Using the sum formula $\frac{n(n+1)}{2}$ with $n=39$:$\frac{39 	imes 40}{2} = 39 	imes 20 = 780$.

The number of points,having both coordinates as integers,that lie in the interior of the triangle with vertices $(0,0)$,$(0,41)$,and $(41,0)$ is:

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