Question

The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of triangle with vertices (0, 0), (0, 21) and (21, 0) is:
A. 133
B. 190
C. 233
D. 105

Answer 1

Hint: For the above question we will first draw the diagram of a triangle with given vertices. Then suppose the point (a, b) that lies inside the triangle and use the condition of two points which lie on the same side of the line.

Complete step-by-step answer:

Let the vertices of the triangle be A (21, 0), B (0, 21) and C (0, 0) then the triangle can be drawn as follows:

Also, let us suppose the (a, b) lies inside of the triangle.

Thus (a, b) lies in the first quadrant from the figure of the triangle.

Therefore, a > 0 and b > 0.

The point (a, b)and C (0, 0) lies on the same side of AB.

Now equation of AB in two point form is as follows:

$y-0=\dfrac{21-0}{0-21}(x-21)$

$y+x-21=0$

For point $\mathrm{C}(0,0)$, we have,

$x+y-21=-21<0$

Therefore,

$a+b-21<0$

$a+b<21$

For $a=1$,

$b<21-1 \Rightarrow b<20: b \in[1,19]$

So, there are a total 19 integral values.

For $a=2$,

$b<21-2 \Rightarrow b<19: b \in[1,18]$

So, there are a total 18 integral values.

Similarly, for $a=19$,

$b<21-19 \Rightarrow b=1$

So, there is only 1 integral value.

Hence, total number of integral points $=19+18+\ldots \ldots+1$ As we know that $1+2+3 \ldots \mathrm{n}=\dfrac{n(n+1)}{2}$

By using the above formula, we get, $19+18+\ldots+1=1+2+3+\ldots+18+19$

$19+18+\ldots+1=\dfrac{19(19+1)}{2}$

$19+18+\ldots+1=\dfrac{19 \times 20}{2}$

$19+18+\ldots+1=19 \times 10$

$19+18+\ldots+1=190$

Therefore, there are a total 190 integral points, which lies inside the triangle

and the correct option is option $\mathrm{B}$.

Therefore, there are a total 190 integral points, which lies inside the triangle and the correct option is option B.

Note: Be careful while writing the equation of AB line as there is a chance that you might make a mistake of sign. Whenever we get this type of question with coordinates (n,0) and (0,n) that cut the axes, we just have to remember that we have to find the sum of terms from 1 to (n-2) to get the total integral points. This can be used to cross check if we have solved it the right way or not.