Question

the number of points required to form 36 line segments no 3 points are collinear is i will mark brainliest who give

Answer 1

Step-by-step explanation:

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Answer 2

$\large\underline\purple{\bold{Solution :- }}$

☆ Let number of points to be required be 'n'.

We know that

$\boxed{ \red{\sf \:^{n}C_r=\dfrac{n!}{(n-r)!\times r!}}}$

$:\implies \sf\boxed{\sf \:^{n}C_2=\dfrac{n!}{(n-2)!\times 2!}}$

$:\implies \sf\boxed{\sf \:^{n}C_2=\dfrac{n(n - 1)(n - 2)!}{(n-2)!\times 2 \times 1}}$

$:\implies \sf\boxed{\sf \:^{n}C_2=\dfrac{n(n - 1)}{2}}$

☆ We know when 2 points are joined, we get one line segment.

☆ So number of line segments having 'n' points in plane is

$:\implies \sf \:^{n}C_2$

☆ According to statement, number of line segments = 36

$:\implies \sf \:^{n}C_2 \: = \: 36$

$:\implies \sf \:\dfrac{n(n - 1)}{2} = 36$

$:\implies \sf \: n(n - 1) = 72$

$:\implies \sf \: n(n - 1) = 9 \times 8$

$:\implies \sf \: on \: comparing \: we \: get$

$:\implies \sf \: n \: = 9$

☆ So, number of points required is 9