The points of intersection of three lines 2X + 3Y - 5 = 0, 5X - 7Y + 2 = 0 and 9X - 5Y - 4= 0

a. form a triangle

b. are on lines perpendicular to each other

c. are on lines parallel to each other

d. are coincident

• The lines are Coincident
  
  Given lines are
  2x+3y-5=0
  5x-7y+2=0
  9x-5y-4=0
  
  If we suppose the three lines form a triangle then area is calculated as
  =>
  |a1 b1 c1|
  |a2 b2 c2|
  |a3 b3 c3|
  
  =>a1((b2*c3)-(b3*c2))-b1((a2*c3)-(a3*c2))+c1((a2*b3)-(a3*b2))
  
  by substituting the values in to the above formula
  =>
  
  |2 3 -5|
  |5 -7 2|
  |9 -5 -4|
  
  =>2((-7*-4)-(-5*2))-3((5*-4)-(9*2))+(-5)((5*-5)-(9*-7))
  =>2(28-(-10))-3(-20-18)+(-5)(-25-(-63))
  =>2(28+10)-3(-38)+(-5)(-25+63)
  =>2(38)+3(38)+(-5)(38)
  since 38 is common we take it as
  =>38(2+3-5) = 0
  
  since the area is zero it represents the lines do not form a triangle, because every triangle has some area
  
  The area Zero is only possible when the lines are coincident
  
  if wanted to verify substitute point (1,1) in the equations it gives all of them to zero, it means that they coincide at the point (1,1).
• 10 years agoHelpfull: Yes(0) No(0)