Given equation, 15x2+31xy+14y2=0 ⇒15x2+10xy+21xy+14y2=0 ⇒5x(3x+2y)+7y(3x+2y)=0 ⇒(3x+2y)(5x+7y)=0 ∴ Two lines have following equation 3x+2y=0 ...(i) 5x+7y=0 ...(ii) Perpendicular distance from (2,3) to the line 3x+2y=0 is given as |
3(2)+2(3)
√9+4
|=
12
√13
Perpendicular distance from (2,3) on line 5x+7y=0 is given as |