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Re: Right triangle PQR is to be constructed in the xy-plane so t [#permalink]
4
-4<=x<=5
6<=y<=16

Let's find out how many possible integer values of x and y we'll have available to choose from

We have
5-(-4)+1=10 x-values to choose from
16-6+1=11 y-values to choose from


Event1: choose an x coordinate for P
Event2: choose a y coordinate for P
Event3: choose an x coordinate for R
Event4: choose a y-coordinate for R
Event5: choose an x-coordinate for Q
Event6: choose a y-coordinate for Q

Possible outcomes of each Event
|Event1|: 10
|Event2|: 11
|Event3|: 9 (b/c we can't choose the same coordinate as P in Event1)
|Event4|: 1 (since PR is parallel to the x-axis we have to choose the same y-coordinate as P in Event2)
|Event5|: 1 (since the right angle occurs at P, Q must be at the same x-coordinate as P in Event1)
|Event6|: 10 (since we can't choose the same y-coordinate as P, we have to chose from the remaining 10)

Notice that the number of outcomes of each event are independent of each other. Regardless of what outcome take place the number of outcomes of each event is the same.

We can use Fundamental Principle of Counting.

Total Outcomes Overall= |Event1|*|Event2|*|Event3|*|Event4|*|Event5|*|Event6| = 10*11*9*1*1*10 = 9900

Final Answer: C
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Re: Right triangle PQR is to be constructed in the xy-plane so t [#permalink]
1
x-axis: P and Q will always have same x coordinate so lets not consider Q in this scenario

P : can have values from -4 to 5. So, P can be placed in any one of the 10 coordinates
R : can also have values from -4 to 5 but P has already taken a point and R cannot take the same position because then it wont be a triangle.
So R can be placed in one of the 9 coordinates.

So in x-axis alone you can have 10*9 bases(PR) of the triangle.

y-axis : P and R will always have same x coordinate so lets not consider R in this scenario

P : can have values from 6 to 16. So, P can be placed in any one of the 11 coordinates
Q : can also have values from 6 to 16 but P has already taken a point and Q cannot take the same position because then it wont be a triangle anymore.
So Q can be placed in one of the 10 coordinates.

So in y-axis alone you can have 11*10 bases(PQ) of the triangle.

Thus total triangle you can have is 9*10*11*10 = 9900.
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Re: Right triangle PQR is to be constructed in the xy-plane so t [#permalink]
1
To approach this, we have to find ways to select each of the coordinates (x,y) of Points P, Q, and R.

Such as P, Q, and R follow the x and y constraints

Anyways, first let's find out the total possible points we can select for both x, and y

for x, we have 10 possible points
for y, we have 11 possible points

Now, Let's check the number of ways Q can take values.

X coordinate of Q = 10C1 ways
Y coordinate of Q = 11C1 ways

Q is now dealt with

For P, the x coordinate is now fixed (as PQR is a right angle triangle and QP is perpendicular to PR, which is parallel to the X axis) because QP is parallel to the y-axis

but we still need to decide a y coordinate -

Therefore,
Y coordinate of P = 10C1 ways (One point is occupied by Q)

Now for R. We know, its y coordinate is fixed as PR is parallel to X axis

Therefore, for its Y coordinate =
Y coordinate of R = 9C1 ways (one point is occupied by P)


Therefore, total combinations =
[(X coordinate of Q = 10C1 ways) * (Y coordinate of Q = 11C1 ways)] * [Y coordinate of P = 10C1 ways] * [Y coordinate of R]
= 10*11*10*9
= 9,900

Thank You

Sincerely
Daksh Kumar
Prep Club for GRE Bot
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