(Q). Consider 5 points in a plane are situated so that no two of the straight lines joining them are parallel, perpendicular, or co-incident. From each point perpendiculars are drawn to all the lines joining the other four points. Determine the maximum number of intersections that these perpendiculars can have?
Solution:-
Consider 5 points A, B, C, D, E. Total ⁵C₂ = 10 lines can be formed by joining these points. Consider a line AB. In this line total 3 perpendiculars can be drawn from remaining 3 points C, D, E.
So total number of perpendiculars
Hence maximum number of points of intersection of these perpendiculars = ³⁰C₂ = 435
but on close observation we found that these 435 points are not distinct means out of 30 perpendicular straight lines not all of them are non concurrent so we have to remove some values out of 435 points. Consider following cases:-
case(i): Total number of triangle that can be formed out of these 5 points = ⁵C₃ = 10
Now consider a triangle △ABC
perpendiculars are drawn from point A, B, C to respective vertices. These perpendiculars intersect at one point known as orthocentre but in calculation of 435 points we have counted number of intersection of these perpendiculars as ³C₂ = 3. But instead of 3 they intersect at one point. So for one triangle we lose 2×10 = 20 points.
Hence we have to subtract 20 points ut of 435 points.
Case(iii):- Consider point A. Remember foour points B, C, D, E ⟹ out of these ⁴C₂ = 6 lines can be drawn.
So from point A 6 perpendiculars will be drawn on these 6 lines but these perpendiculars are concurrent i.e. they interect at one point but in calculation of 435 intersection point for these 6 perpendiculars we have counted point of intersection as ⁶C₂ = 15 but actually they intersect at only 1 point so 14 intersection points are counted extrafor point A. So overall 14×5 = 70 points are counted extra which we have to remove from original calculation.
Case(ii): Consider line AB
A case may arise when perpendiculars drawn on AB from C, D & E are parallel to each other i.e. they do not intersect but we have counted their number of point of intersection as ³C₂ = 3
So we have to subtract 3×10 = 30 from original calculation
↓
10 lines from
points A, B, C, D, E
& No other case is possible
Hence maximum number of points of intersection of perpendiculars = 435 – case(i) – case(ii) – case(iii)
= 435 – 20 – 70 – 30
= 315 Answer