(3)/(x)-(1)/(y)+9=0;quad (2)/(x)+(3)/(y)=5quad (x!=0,y!=0)

(3) / (x) - (1) / (y) + 9 = 0 (2) / (x) + (3) / (y) = 5, (! = 0, y! = 0)

( 3 ) /(x) - (1 ) /( y) + 9 = 0 , (2)/(x) + ( 3)/( y) = 5 ( x ne 0 , y ne 0 )

Solve by cross-multiplication method. (i) 8x-3y=12,5x=2y +7 (ii) 6x+ 7y -11 =0, 5x+ 2y =13 (iii) (2)/(x)+(3)/(y) =5, (3)/(x) -(1)/(y) +9=0

( 5)/(x)- ( 3) /( y ) = 1 , ( 3) /( 2 x ) + ( 2) /( 3y ) = 5 ( x ne 0 , y ne 0 )

Solve by cross-multiplication method (2)/( x) + ( 3)/( y ) = 5, ( 3)/( x ) - (1)/( y ) + 9 = 0

2(2x+y+5)+3(x-3y-1)=0 2x-3y+1=0

Solution of D.E (dy)/(dx)=(2x+5y)/(2y-5x+3) is,if (y(0)=0) (1) x^(2)-y^(2)+5xy-3y=0 (2) x^(2)+y^(2)+5xy-3y=0 (3) x^(2)-y^(2)+5xy+3y=0 (4) x^(2)-y^(2)-5xy-3y=0

The locus of a point "P" ,if the join of the points (2,3) and (-1,5) subtends right angle at "P" is x^(2)+y^(2)-x-8y+13=0 x^(2)-y^(2)-x+8y+3=0 x^(2)+y^(2)-4x-4y=0,(x,y)!=(0,4)&(4,0) x^(2)+y^(2)-x-8y+13=0,(x,y)!=(2,3)&(-1,5)

If one of the diagonals of a square is along the line x=2y and one of its vertices is (3, 0), then its sides through this vertex are given by the equations (A) y-3x+9=0, 3y+x-3=0 (B) y+3x+9=0, 3y+x-3=0 (C) y-3x+9=0, 3y-x+3=0 (D) y-3x+9=0, 3y+x+9=0