[" (i) "x-3y-3=0],[3x-9y-2=0],[" (iti) "3x-5y=20],[6x-10y=40]

6x-3y=-10,3x+5y-8=0

6x-3y+10=0,2x-y+9=0

6x-3y+10=0 2x-y+9=0

The lines x-y-6=0, 4x-3y-20=0 and 6x+5y+8=0 are:

2x-y-5=0; x+3y-9=0

Solve using the method of substitution. ( i) 2x-3y =7,5x +y=9 (ii) 1.5x+ 0.1 y =6.2,3x -0.4y=11.2 (iii) 10% of x +20% of y =24, 3x-y=20 (iv) sqrt2x -sqrt 3y =1, sqrt 3x -sqrt 8y =0

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

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

Four lines x+3y-10=0, x+3y-20=0, 3x-y + 5 = 0 and 3x - y - 5 = 0 form a figure which is.

Prove that the following lines are concurrent. (i)5x-3y=1 , 2x+3y=23 , 42x+21y=257 (ii) 2x+3y-4=0 , x-5y+7=0 , 6x-17y+24=0