Consider the set of all triangles OPQ where O | vedclass

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(C) The coordinates of points $P$ and $Q$ lie on the line $5x + y = 99$. Since $x$ and $y$ are non-negative integers,$x$ can take values from $0$ to $

Vedclass · Accepted Answer

$90$

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(C) The coordinates of points $P$ and $Q$ lie on the line $5x + y = 99$. Since $x$ and $y$ are non-negative integers,$x$ can take values from $0$ to $19$ (as $5(19) + 4 = 99$ and $5(20) = 100 > 99$).Let $P = (x_1, 99 - 5x_1)$ and $Q = (x_2, 99 - 5x_2)$,where $x_1, x_2 \in \{0, 1, 2, \dots, 19\}$ and $x_1 
eq x_2$.The area of triangle $OPQ$ with vertices $(0,0)$,$(x_1, y_1)$,and $(x_2, y_2)$ is given by $\Delta = \frac{1}{2} |x_1 y_2 - x_2 y_1|$.Substituting $y_1 = 99 - 5x_1$ and $y_2 = 99 - 5x_2$:$\Delta = \frac{1}{2} |x_1(99 - 5x_2) - x_2(99 - 5x_1)|$$\Delta = \frac{1}{2} |99x_1 - 5x_1 x_2 - 99x_2 + 5x_1 x_2|$$\Delta = \frac{99}{2} |x_1 - x_2|$.For $\Delta$ to be an integer,$|x_1 - x_2|$ must be an even number (since $99$ is odd).Thus,$x_1$ and $x_2$ must have the same parity (both even or both odd).In the set $\{0, 1, 2, \dots, 19\}$,there are $10$ even numbers and $10$ odd numbers.The number of ways to choose two distinct even numbers is $^{10}C_2 = \frac{10 	imes 9}{2} = 45$.The number of ways to choose two distinct odd numbers is $^{10}C_2 = \frac{10 	imes 9}{2} = 45$.Total number of such triangles = $45 + 45 = 90$.

Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$,$Q$ are distinct points in the plane with non-negative integral coordinates $(x, y)$ such that $5x + y = 99$. The number of such distinct triangles whose area is a positive integer is:

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