Approach #1:

There are two types of triangles possible:
With two vertices on the line with 4 points and the third vertex on the line with 5 points --> \(C^2_4*C^1_5=30\);
With two vertices on the line with 5 points and the third vertex on the line with 4 points --> \(C^2_5*C^1_4=40\);

Total: \(30+40=70\).

Answer: D.


Approach #2:

All different 3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear.
\(C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70\) (where \(C^3_4\) and \(C^3_4\) are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).

Answer: D.

Hope it's clear.
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Not that fun this question I wonder how often I would see such a question on test day.

Solution:

* * * * * (line B)

* * * * (line A)

Choose 2 points from line A and a point from line B: = 4!/2!2! * 5!/1!4! = 30
Choose 1 point from line A and 2 points from line B: = 4!/1!3! * 5!/2!3! = 4 * 10 = 40

Combine possiblities: 30 + 40 = 70

I really like your 2nd approach

Combination problems are located at search.php?search_id=tag&tag_id=52

We have two scenarios:

The first scenario is that the 1 vertex is on the first line and the 2 other vertices are on the second line.
So the number of ways to create such a triangle is:

4C1 x 5C2 = 4 x (5 x 4)/2 = 40

The second scenario is that 2 vertices are on the first line and the 1 vertex is on the second line. So the number of ways to create such a triangle is:

4C2 x 5C1 = (4 x 3)/2 x 5 = 6 x 5 = 30

Thus, the total number of ways to create the triangle is 40 + 30 = 70.

Answer: D
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