
The number of integral points (integral means both the coordinates should be integer) exactly in the interior of the triangle with vertices \[\left( {0,0} \right)\], \[\left( {0,21} \right)\] and \[\left( {21,0} \right)\] is
A. 133
B. 190
C. 233
D. 105
Answer
587.4k+ views
Hint: In this problem, first we need to obtain the equation of the line joining the points \[\left( {21,0} \right)\] and \[\left( {0,21} \right)\]. Then, find out the number of integral points which lies exactly in the interior of the triangle.
Complete step by step answer:
Let, the vertices of the triangle be \[P\left( {21,0} \right),Q\left( {0,21} \right)\,\,{\text{and}}\,\,O\left( {0,0} \right)\].
Consider, a point \[\left( {a,b} \right)\]. The point \[\left( {a,b} \right)\] is in the interior of the triangle if it lies in first quadrant.
Therefore,
\[
a > 0 \\
b > 0 \\
\]
The equation of the line joining the points \[P\] and \[Q\] is as follows:
\[
\,\,\,\,\,x + y = 21 \\
\Rightarrow x + y - 21 = 0 \\
\]
The point \[\left( {a,b} \right)\] lies below the line \[PQ\], therefore,
\[
a + b - 21 < 0 \\
\Rightarrow a + b < 21 \\
\]
For \[a = 1:b < 20 \in \left[ {1,19} \right]\] total 19 values of \[b\].
For \[a = 2:b < 19 \in \left[ {1,18} \right]\] total 18 values of \[b\].
………………………….
………………………….
Similarly, For \[a = 19:b < 2\] total 1 value of \[b\].
Thus, the number of integral points is obtained as follows:
\[
{\text{Number of integral points = 19 + 18 + 17 + }} \ldots + 1 \\
{\text{Number of integral points}} = \dfrac{{19 \times \left( {19 + 1} \right)}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{sum of }}n{\text{ natural numbers}} = \dfrac{{n\left( {n + 1} \right)}}{2}} \right] \\
{\text{Number of integral points}} = \dfrac{{19 \times 20}}{2} \\
{\text{Number of integral points}} = 190 \\
\]
So, the correct answer is “Option B”.
Note: We can find the sum of n terms using the formula of the arithmetic progression also. The formula for the sum of \[n\] numbers in arithmetic progression is \[\dfrac{{n\left( {2a + \left( {n - 1} \right)d} \right)}}{2}\,\,\,{\text{or}}\,\,\,\dfrac{{n\left( {a + l} \right)}}{2}\], here a is first term, l is last term and d is common difference.
Complete step by step answer:
Let, the vertices of the triangle be \[P\left( {21,0} \right),Q\left( {0,21} \right)\,\,{\text{and}}\,\,O\left( {0,0} \right)\].
Consider, a point \[\left( {a,b} \right)\]. The point \[\left( {a,b} \right)\] is in the interior of the triangle if it lies in first quadrant.
Therefore,
\[
a > 0 \\
b > 0 \\
\]
The equation of the line joining the points \[P\] and \[Q\] is as follows:
\[
\,\,\,\,\,x + y = 21 \\
\Rightarrow x + y - 21 = 0 \\
\]
The point \[\left( {a,b} \right)\] lies below the line \[PQ\], therefore,
\[
a + b - 21 < 0 \\
\Rightarrow a + b < 21 \\
\]
For \[a = 1:b < 20 \in \left[ {1,19} \right]\] total 19 values of \[b\].
For \[a = 2:b < 19 \in \left[ {1,18} \right]\] total 18 values of \[b\].
………………………….
………………………….
Similarly, For \[a = 19:b < 2\] total 1 value of \[b\].
Thus, the number of integral points is obtained as follows:
\[
{\text{Number of integral points = 19 + 18 + 17 + }} \ldots + 1 \\
{\text{Number of integral points}} = \dfrac{{19 \times \left( {19 + 1} \right)}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{sum of }}n{\text{ natural numbers}} = \dfrac{{n\left( {n + 1} \right)}}{2}} \right] \\
{\text{Number of integral points}} = \dfrac{{19 \times 20}}{2} \\
{\text{Number of integral points}} = 190 \\
\]
So, the correct answer is “Option B”.
Note: We can find the sum of n terms using the formula of the arithmetic progression also. The formula for the sum of \[n\] numbers in arithmetic progression is \[\dfrac{{n\left( {2a + \left( {n - 1} \right)d} \right)}}{2}\,\,\,{\text{or}}\,\,\,\dfrac{{n\left( {a + l} \right)}}{2}\], here a is first term, l is last term and d is common difference.
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