Let $X=\left\{(x,y)\in\mathbb{Z}\times\mathbb{Z}:\dfrac{x^2}{8}+\dfrac{y^2}{20}<1\text{ and }y^2<5x\right\}$. Three distinct points $P$, $Q$ and $R$ are randomly chosen from $X$. Then the probability that $P$, $Q$ and $R$ form a triangle whose area is a positive integer, is
Step-by-Step Solution
Key Concept: For points (1,y₁),(1,y₂),(2,y₃): area = |y₁−y₂|/2; integer iff y₁,y₂ same parity
$y^2<5x$ requires $x>0$, so $x\in\{1,2\}$.
$x=1$: $y^2<5$, $y\in\{-2,-1,0,1,2\}$ (5 points; ellipse check: $1/8+y^2/20<1$ ✓ for all).
$x=2$: $y^2<10$, $y\in\{-3,-2,-1,0,1,2,3\}$ (7 points; $4/8+y^2/20<1\Rightarrow y^2<10$ ✓).
$|X|=12$. Total triples $=\binom{12}{3}=220$.
**Degenerate (collinear):** All from $x=1$: $\binom{5}{3}=10$. All from $x=2$: $\binom{7}{3}=35$. Total degenerate: 45.
**Non-degenerate (mixed):** 175 triples. For these, area$=(1/2)|$det$|$.
*2 from $x=1$, 1 from $x=2$:* det $=y_1-y_2$. Integer area iff $|y_1-y_2|$ even iff $y_1\equiv y_2\pmod{2}$.
Same-parity pairs from $\{-2,-1,0,1,2\}$: even$\{-2,0,2\}$→$\binom{3}{2}=3$; odd$\{-1,1\}$→1. Total 4 pairs $\times$ 7 choices $=28$.
*1 from $x=1$, 2 from $x=2$:* det $=y_3-y_2$. Integer area iff $y_2\equiv y_3\pmod{2}$.
Same-parity pairs from $\{-3,-2,-1,0,1,2,3\}$: even 3, odd 4. $\binom{3}{2}+\binom{4}{2}=3+6=9$ pairs $\times$ 5$=45$.
Total with positive integer area: $28+45=73$. Probability $=73/220$.
Correct Answer: B