Question

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

Answer 1

Answer:

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.

$S_{n}=\frac{n}{2}*[{\text{first term}}+{\text{last term}}]\\\\S_{39}=\frac{39}{2}[1+39]=39 *20=780$

Number of integral  points in the interior of triangle with vertices (0,0),(0,41),(41,0)

is​ = 780

Answer 2

The point on the hypotenuse is (1,40)

(

1

,

40

)

. The number of points inside the triangle are 39

39

, that is, {1,2…,39}

{

1

,

2

…

,

39

}

.

Similarly number of integral points on =2

x

=

2

are 38

38

, that is {1,2,…,38}

{

1

,

2

,

…

,

38

}

.

So basically you have to sum up 39+38+⋯+1=40.392=780.