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It is given that two vertices of square are (0,0) and (4,2).
Now the problem is that you haven't given that whether these two vertices are adjacent vertices or opposite vertices of the square.
1. By Supposing that these two are adjacent vertices of Square
The third vertex will be at (-4,2) which lies in third quadrant.
Suppose the coordinate of fourth vertex be (x,y).
Mid point of line joining (4,2) and (-4,2) is{ [4+(-4)]/2,(2+2)/2} is (0,2).
Mid point of line joining (x,y) and (0,0) is (x/2,y/2).
Since diagonals of square bisect each other,
∵ x/2=0
⇒x=0
and
y/2=2
⇒y=4
So, The Coordinate of fourth vertex is (0,4).
Now coming back to second condition if these are two opposite vertex of Square.
Let the third coordinate be (a,b).
Length of diagonal=
Now,let side of Square be A.
Then length of Diagonal of square =√2 A
⇒√2 A=2√5
⇒A =√10
As third vertex is (a,b).
Using distance formula
a² + b²=10 -------------(1)
(a-4)²+ (b-2)²=10 --------------(2)
Solving expression (1) and (2), we get
⇒a²+ b²=(a-4)² +(b-2)²
⇒2a + b =5
⇒b=5-2a
Putting the value of b in (1),we get
⇒a² +(5-2a)²=10
⇒a²+25+4a²-20a =10
⇒5a²-20a+15=0
⇒a² - 4a + 3=0
Splitting the middle term,we get
⇒(a-3)(a-1)=0
⇒a=3 ∧ a=1
we get b=5-2×1=3 and b=5-2×3=5-6=-1
So,the other vertex are (1,3) and(3,-1).
First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of
possible passwords.
Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of
possible passwords.
So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is
Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have
As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:
Where x can be any of the 36 characters.
So, we have 25 cases with 4 vacant slots, leading to a probability of
Finally, you can compute the probability of the union using the formula
Since we already computed all these quantities.