Out of 18 points in a plane,no three are in t | vedclass

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(C) Given $18$ points in a plane,where $5$ points are collinear.$(i)$ Number of straight lines:Total lines possible from $18$ points is $^{18}C_2$.Sin

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$(i) 144, (ii) 806$

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(C) Given $18$ points in a plane,where $5$ points are collinear.$(i)$ Number of straight lines:Total lines possible from $18$ points is $^{18}C_2$.Since $5$ points are collinear,they form only $1$ line instead of $^5C_2$ lines.Number of lines $= ^{18}C_2 - ^5C_2 + 1 = \frac{18 	imes 17}{2} - \frac{5 	imes 4}{2} + 1 = 153 - 10 + 1 = 144$.$(ii)$ Number of triangles:Total triangles possible from $18$ points is $^{18}C_3$.Since $5$ points are collinear,they do not form any triangle.Number of triangles $= ^{18}C_3 - ^5C_3 = \frac{18 	imes 17 	imes 16}{3 	imes 2 	imes 1} - \frac{5 	imes 4 	imes 3}{3 	imes 2 	imes 1} = 816 - 10 = 806$.

Out of $18$ points in a plane,no three are in the same straight line except five points which are collinear. The number of $(i)$ straight lines and $(ii)$ triangles which can be formed by joining them is:

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