A point $(x, y)$ with integer coordinates is randomly selected such that $0 \le x \le 8$ and $0 \le y \le 4$. what is the probability that $x + y \le 4$? express your answer as a common fraction.

Step-by-step explanation:

A point (x, y) with integer coordinates is randomly selected such that

The possible pairs of (x,y) are:

(0,0),(0,1),(0,2),(0,3),(0,4)

(1,0),(1,1),(1,2),(1,3),(1,4)

(2,0),(2,1),(2,2),(2,3),(2,4)

(3,0),(3,1),(3,2),(3,3),(3,4)

(4,0),(4,1),(4,2),(4,3),(4,4)

(5,0),(5,1),(5,2),(5,3),(5,4)

(6,0),(6,1),(6,2),(6,3),(6,4)

(7,0),(7,1),(7,2),(7,3),(7,4)

(8,0),(8,1),(8,2),(8,3),(8,4)

The Total Possible Outcomes n(S)= 45

The pair (x, y) that satisfies the given condition (say event A: ) are:

n(A)=15

Therefore:

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

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