Let X= (x, y) Z Z: x^2 8 +y^2 20 <1 . and .y^2<5 x . Three distin… - Vidyadip

MCQ

Let $X=\left\{(x, y) \in \mathbb{Z} \times \mathbb{Z}: \frac{x^2}{8}+\frac{y^2}{20}<1\right.$ and $\left.y^2<5 x\right\}$. 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

A
$\frac{71}{220}$
✓
$\frac{73}{220}$
C
$\frac{79}{220}$
D
$\frac{83}{220}$

✓

Answer

Correct option: B.

$\frac{73}{220}$

b
$\frac{x^2}{8}+\frac{y^2}{20}<1 y^2<5 x$

Solving corresponding equations

$\frac{x^2}{8}+\frac{y^2}{20}=1 y^2=5 x$

$x=2$

$y= \pm \sqrt{10}$

$X=\{(1,1),(1,0),(1,-1),(1,2),(1,-2),(2,3),(2,2),(2,1),(2,0),(2,-1),(2,-2),(2,-3)\}$

(image)

Let $\mathrm{S}$ be the sample space $\mathrm{E}$ be the event $\mathrm{n}(\mathrm{S})={ }^{12} \mathrm{C}_3$

For $E$

Selecting $3$ points in which $2$ points are either or $\mathrm{x}=1$ $\mathrm{x}=2$ but distance $\mathrm{b} / \mathrm{w}$ then is even

Triangles with base $2$ :

$=3 \times 7+5 \times 5=46$

Triangles with base $4$ :

$=1 \times 7+3 \times 5=22$

Triangles with base $6$ :

$=1 \times 5=5$

$P(E)=\frac{46+22+5}{{ }^{12} C_3}=\frac{73}{220}$

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