Algebra
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Q.

The right angled triangle $PQR$ is to be constructed in the $xy$-plane, so that the right angle is at $P$ and $PR$ is parallel to the $x$-axis. The $x$ and $y$ coordinates of $P,Q$ and $R$ are to be integers that satisfy the inequality $−4≤x≤5$ and $6≤y≤16$.

How many different triangles with these properties could be constructed?

 A.

1,100

 B.

12,100

 C.

10,000

 D.

9,900

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Solution:
Option(D) is correct

$X$ coordinate has 10 possible values between -4 and 5 

$Y$ coordinate has 11 possible values between 6 and 16 

1. Fix $P$ first, $P$ can have 10 values across $X$ axis and 11 across $Y$ axis. So, that's $11×10=110$ 

2. Since $PR$ is parallel to $X$ axis, $R$ is on the same line as $P$. So, $R  (=10−1)$ values left along $X$ axis.

So far $110×9=990 $

3. Since $PQR$ is a right angle triangle and the right angle is at $P, PQ$ is parallel to $Y$ axis. So, $Q$ has $10( =11−1)$ values left.

So, $990×10=9,900$

Edit: For an alternative solution making use of the combination, check comment by Kavya.


(4) Comment(s)


Sam
 ()

Fixing P: P can take any value from 10(for x-axis) and 11(for y-axis) = 10*11= 110

Fix R: R can have only 1 Y- value (same as that of p) and x value as that of y =1*(10-1)=9

Fix Q: Q will have same x value as p and different y value from P & R =1*(11-1) = 10

Final ans =110*9*10 =9900



Kavya
 ()

There are 10 different values for $x$ and 11 values for $y$.

So, the number of triangles that can be formed taking 3($x$ values) and 3,($y$ values) would be ${^{10}C_3} \times {^{11}C_3} \times \left(\dfrac{1}{2}\right) = 9,900$

It is multiplied by $\frac{1}{2}$ to account for the repetitions.


Vivek
 ()

I think your solution counts every possible triangle with the given set of points. Why you do $1/2$ is clearly not understood. However you get the right answer anyways $:)$

Shubham Kundu
 ()

this is totally wrong, the correct formula is mn(m-1)(n-1), explanation is given in the main solution