6 points are place on the line a, 4 points are placed on the line b. How many triangles is it possible to form such that their verticies will be the given points, if a ∥b?

96

Step-by-step explanation:

Ok, lines a and b are parallel.

We can separate this problem in two cases:

Case 1: 2 vertex in line a, and one vertex in line b.

Here we use the relation:

"In a group of N elements, the total combinations of sets of K elements is given by"

Here, the total number of points in the line is N, and K is the ones that we select to make the vertices of the triangle.

Then if we have two vertices in line a, we have:

N = 6, K = 2

And the other vertex can be on any of the four points on the line b, so the total number of triangles is:

C = 15*4 = 60.

But we still have the case 2, where we have 2 vertices on line b, and one on line a.

First, the combination for the two vertices in line b is:

We use N = 4 and K = 2.

And the other vertice of the triangle can be on any of the 6 points in line a, so the total number of triangles that we can make in this case is:

C = 6*6 = 36

Then, putting together the two cases, we have a total of:

60 + 36 = 96 different triangles

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.