Official answer is "C" - 9900

This was a hard question for me.

Anyway, I took the following approach to solve this.

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When y=6

We know -4<=x<=+5, there are 10 integral points.

The points are: {-4,6},{-3,6},{-2,6},{-1,6},{0,6},{1,6},{2,6},{3,6},{4,6},{5,6}

So, number of distinct line segments that can be formed are 10C2=45.

And these line segments PR as it should be parallel to x-axis.

With every line segment there can be 20 triangles:

Let's see how:

consider line segment with points: P={-4,6} and R={1,6}

We know there are 10 points above P that can serve us as Q. Q should always be

vertically above or below point P because P is the right angle.

So Q can be: {-4,7},{-4,8},{-4,9},{-4,10},{-4,11},{-4,12},{-4,13},{-4,14},{-4,15},{-4,16}

See for yourself that now: for just one segment PQ we have 10 triangles.

PQR.

However, we can flip the points on the same line segment PR, so that P={1,6} and R={-4,6}

We know there are 10 points above P that can serve us as Q. Q should always be

vertically above or below point P because P is the right angle.

So in this case Q can be:

{1,7},{1,8},{1,9},{1,10},{1,11},{1,12},{1,13},{1,14},{1,15},{1,16}

See for yourself that for just one segment PQ we have another 10 triangles.

So, a total of 20 distinct triangles for just one line segment.

thus: a total of 20 * 10C2 = 20 * 45 = 900 triangles for all line segments where PR

is on y=6.

Remember, we were talking only about PR that lies on y=6 line.

There are 11 such lines between y=6 and y=16 i.e. 16-6+1.

Thus; a total of 900 * 11 = 9900 triangles can be formed with given conditions.

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~fluke

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~fluke

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