it will screw you....

There are $10$ possible $x-coordinates$ and $11$ possible $y-coordinates$ for point $Q$ . Point $R$ cannot have the same $x$ and $y-coordinates$ with that of $Q$ . Thus, point $R$ has $9$ possible $x-coordinates$ and $10$ possible $y-coordinates$ . Point $P$ will have the same $x-coordinate$ with that of $Q$ and the same $y-coordinate$ with that of $R$ .

So, the number of triangles that may be formed = $(10)(11)(9)(10)=9900$ .

Edit: Added image.

Using the given range of possible coordinates for the ${x}$ and ${y}$ values of each point, there are 100 total possible points with 10 ${x}$ values and 10 ${y}$ values. Choosing any two points Q and R, $({x}_{q},{y}_{q})$ and $({x}_{r},{y}_{r})$ respectively gives only two points possible for P, $({x}_{r},{y}_{q})$ and $({x}_{q},{y}_{r})$ , such that there is a right angle at P. With 100 potential cooridinates, the number of possible combinations of Q and R is 100*99=9900, and for each of these there are two possible locations of P. However, each of these scenarios is repeated twice, as when Q and R are switched the same triangle is made. Since there are two trianlges in each scenario but each of the 9900 scenarios occurs twice there are only 9900 possible triangles

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