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There are 23 straight lines in a plane, of which 9 and 7 passes through points A and B, respectively. In addition, no line crosses both points A and B, no two lines are parallel, and no three lines intersect a single point. Determine how many points the straight lines intersect.

Option: 1

200 
 


Option: 2

198 


Option: 3

218

 


Option: 4

250 


Answers (1)

best_answer

Given that,

There are 23 straight lines in a plane, of which 9 and 7 pass through points A and B.

There is no line that crosses both points A and B, no two lines are parallel, and no three lines intersect a single point.

Let N represent the overall number of point intersections.

First, the formula ^{n}C_{2} can be used to determine the number of places of the intersection of 23 non-parallel lines supplied, each of which intersects at a different location.

Thus,

\mathrm{\begin{aligned} & N={ }^{23} C_2 \\ & N=\frac{23 !}{2 ! 21 !} \\ & N=\frac{22 \times 23}{2} \\ & N=253 \end{aligned}}

From the given, 9 of those lines pass through point A in accordance with the conditions stated in the question. This indicates that there is only one point at which all 9 lines intersect. Therefore, we will eliminate the intersection of 9 lines and add just 1 for point A. This is given by,

\mathrm{\begin{aligned} & N=253-{ }^9 C_2+1 \\ & N=253-\frac{9 !}{2 ! 7 !}+1 \\ & N=253-\frac{8 \times 9}{2}+1 \\ & N=218 \end{aligned}}

Additionally, 7 additional lines pass through point B (given that no line passes through nd B). both A aTherefore, we will calculate these lines in the same way that we did for the case above. The number of points of intersection will be as follows once these situations have also been eliminated:

\mathrm{\begin{aligned} &N=218-{ }^7 C_2+1\\ &N=218-\frac{7 !}{2 ! 5 !}+1\\ &N=218-\frac{7 \times 6}{2}+1\\ &N=198 \end{aligned}}

Therefore, the number of lines the straight lines intersect is 198.

 

 

 

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