The number of integral points (integral point | vedclass

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(B) The vertices of the triangle are $(0, 0)$,$(0, 21)$,and $(21, 0)$.The interior points $(x, y)$ must satisfy $x > 0$,$y > 0$,and $x + y < 21$.For a

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$190$

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(B) The vertices of the triangle are $(0, 0)$,$(0, 21)$,and $(21, 0)$.The interior points $(x, y)$ must satisfy $x > 0$,$y > 0$,and $x + y For a fixed $x$,$y$ can take integer values from $1$ to $21 - x - 1 = 20 - x$.Since $y > 0$,we must have $20 - x \geq 1$,so $x$ can range from $1$ to $19$.For each $x \in \{1, 2, \dots, 19\}$,the number of possible values for $y$ is $20 - x$.The total number of integral points is $\sum_{x=1}^{19} (20 - x) = 19 + 18 + \dots + 1$.Using the sum formula $\frac{n(n+1)}{2}$ with $n = 19$,we get $\frac{19 	imes 20}{2} = 190$.

The number of integral points (integral point means both the coordinates should be integers) exactly in the interior of the triangle with vertices $(0, 0)$,$(0, 21)$,and $(21, 0)$ is:

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