The number of points having both coordinates as integers
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Answer:
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We have to find those integral points which lie in the interior of triangle with vertices (0,0),(0,41),(41,0) .
We will proceed in the following manner
1. We don't have to consider those points, which lie on the line joining (41,0) and (0,41).Equation of line in slope form with given points is
x + y=41
2. We don't have to consider points on , x axis as well as y axis.
So, restrictions are
that the points shouldn't lie on
1. Line ,x+y =41
2. (x,0)≠[(0,0),(1,0),.......(41,0)]
3. (0,y)≠[(0,0),(0,1),(0,2),.......(0,41)]
As, these three points , (0,0) , (41,0) and (0,41) are not collinear , these points when joined ,will form a right triangle.
Constraints on right triangle to determine number of points in the interior of triangle
x +y <41
x>0, y>0, and x and y must be integers.
Number of points on X axis which we do not count = 42, so if we remove 0, and 41 , number of integral points =40
Proceed from X axis in upward direction that is in first quadrant from 39 points to 1 point,then
Total number of points = 39 +38+37+36+35+34+33+...........+1
which is an airthmetic progression.
\begin{lgathered}S_{n}=\frac{n}{2}*[{\text{first term}}+{\text{last term}}]\\\\S_{39}=\frac{39}{2}[1+39]=39 *20=780\end{lgathered}
S
n
=
2
n
∗[first term+last term]
S
39
=
2
39
[1+39]=39∗20=780
Number of integral points in the interior of triangle with vertices (0,0),(0,41),(41,0)
is = 780
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