Find the set of all points (x, y) such that the area of the triangle with vertices (0, 0), (6, 4) and (x, y) is 4.
(x, y) lies on the circle (y - 6)2 + (x - 4)2 = 16
(x, y) satisfies 6y - 4x = 8 or 6y - 4x = -8
There is no (x, y) satisfying the requirement.
(x, y) satisfies 6y - 4x = 4 or 6y - 4x = -4
(x, y) satisfies 6y - 4x = 8

• area of a triangle={x1(y2-y3)+x2(y3-y1)+x3(y1-y2)}/2
  4={0(y-4)+x(4-0)+6(0-y)}/2
  4x-6y=8 [ans]
  
• 10 years agoHelpfull: Yes(4) No(2)
• Soltn of the eqn of d crcle & d strght lne is (10,8)
  SO this is the point,(10,8)
• 10 years agoHelpfull: Yes(1) No(6)
• ans: (x,y) satisfies 6y - 4x = 8
• 10 years agoHelpfull: Yes(1) No(0)
• find the distance for (x,y) and (6,4)
  (x1-y1)^2+(x2-y2)^2 then get (x,y)
  6y-4x=8 values are (10,8)by use of eqn of cirlce
• 9 years agoHelpfull: Yes(1) No(0)