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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say line A with 4 points and line B with 5 points. and triangle is PQR.
If P is on line A and QR on line B we can have 4C1*5C2=40 triangles.
If P is on line B and QR on line A we can have 4C2*5C1=30 triangles.
Total =70 triangles.

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

See Image below for what my scratch paper looked like on this one

I really like your 2nd approach

Thanks Bunuel. That was great. Do you by any chance have other problems similar to this that you know of?

Combination problems are located at search.php?search_id=tag&tag_id=52
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Attached is a visual that should help. Bunuel's method is the most elegant, but here is an alternative strategy if the other solution is not clear.

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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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1st line: 4 points

2nd line: 5 points

If we choose 1 point on the first line (4C1) and two points from the second line (5C2), we have 4C1 * 5C2 = 40 triangles.

Now, we can choose 2 points on the first line (4C2) and 1 point on the second line (5C1) to get 4C2 * 5C1 = 6 * 5 = 30 triangles

40 + 30 = 70 triangles

Answer is D.
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Out of all the Unique Triangles Possible, every single one of them will have 2 Collinear Points on either the:

-Parallel Line with 5 Dots
or
-Parallel Line with 4 Dots

the 3rd Vertex of the Triangle will have to be on the OTHER Parallel Line in order to create a Triangle with positive area


Case 1: We place the Base on the Bottom Parallel Line with 5 Points

(1st)We need to figure out how many Unique Pairs of 2 Points we can pick out of the 5 Points available on this line.

"5 choose 2" = 5!/2!3! = 10 Unique Combinations

AND

(2nd)For each of these 10 unique combinations, there are 4 Points on the Parallel Line on Top which can combine with the given Pair of Points to create a Triangle with positive area. How many ways can we choose 1 for each of the 10 unique combinations:

"4 choose 1" = 4 ways

(10) * (4) = 40 Unique Triangles in which the "Base" is on the Bottom Parallel Line with 5 Dots


Case 2: We place the Base on the Top Parallel Line with 4 Dots.

Same Logic as above, except the Numbers change:

(1st) "4 choose 2" * "5 choose 1" = [4!/2!2!] * 5 = [6] * 5 = 30 Unique Triangles in which the "Base" is on the Top Parallel Line with 4 Dots


40 Triangles from Case 1 + 30 Triangles from Case 2 =

70 Possible Triangles
(D)

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