Question

: There are 12 points in a plane. If 4 of them are on
a straight line and no other three points are on
a straight line, then find the difference between the
number of triangles and the number of straight lines
that can be formed using these points.​

Answer 1

Answer:

Step-by-step explanation:

(a) We know passing through two points in a plane we can draw only one line, i.e., we require to select any two points from the given 12 points which is possible in .12C2 ways.

∴ The number of different straight lines that can be formed by joining the given 12 points.

.12C212×111×2=66.

(b) Given, out of the 12 points, 4 points are collinear.

We know that collinear points form only one line.

∴ These four points when they are not collinear will actually form .4C2 lines, which are not forming here.

∴ The number of the required lines =.12C2−.4C2+1=66−6+1=61.

(c) We know, by joining three non-collinear points a triangle forms.

∴ Three points can be selected from 12 points in .12C3 ways.

∴ The required number of triangles =.12C3=12×11×101×2×3=220.

(d) Given 5 points are collinear.

⇒.5C3 triangles will not form.

∴ The required number of triangles =.12C3−.5C3=220−10=210.

(d) Given 5 points are collinear.

⇒.5C3 triangles will not form.

∴ The required number of triangles =.12C3−.5C3=220−10=210.

Hope you understood :D