Which is a counterexample for the conditional statement shown? if three distinct points lie in a plane, then the points can form a triangle. the points form a different polygon. the points form a circle. the points form an open figure. the points form a line.

  The points form a line

Step-by-step explanation:

Three points on a line will form a triangle with zero area. Many authors exclude such a triangle from the definition of a triangle. Hence your counterexample will be three points on the same line.

Comments on the definition of a triangle

Where a, b, c are the side lengths, ...

If your version of the triangle inequality is ...

then three collinear points will not form a triangle.

___

If your version of the triangle inequality is ...

then three collinear points will form a triangle.

Comment on the problem statement

Three distinct points will always lie in a plane. If they are not collinear, they define a unique plane. Since the condition is always true, the problem amounts to finding a counterexample for the conclusion. As we see in the discussion above, whether you can do that or not depends on how you define a triangle.

Using the second definition of a triangle, no counterexample exists: three points always define a triangle.

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i