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`

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

Given the four lines with the equations x+2y-3=0 , 3x+4y-7=0 , 2x+3y-4=0 , 4x+5y-6=0 , then

Given the four lines with the equations x+2y-3=0 , 3x+4y-7=0 , 2x+3y-4=0 , 4x+5y-6=0 , then

The lines 3x-4y=9 and y=0 meet at :

Find the solution of the equations 8x + 5y - 9 = 0 and 3x + 2y - 4 = 0 .

The area bounded by the curve x=3y^2-9 and the line x=0,y=0 and y=1 is

Given the four lines with equations x+2y-3=0, 3x+4y-7=0, 2x+3y-4=0 and 4x+5y-6=0 , then:

Write the nature of the equations 6x-2y+9 = 0 and 3x - y +12 = 0

Two straight lines u=0a n dv=0 pass through the origin and the angle between them is tan^(-1)(7/9) . If the ratio of the slope of v=0 and u=0 is 9/2 , then their equations are (a) y+3x=0 and 3y+2x=0 (b) 2y+3x=0 and 3y+x=0 (c) 2y=3x and 3y=x (d) y=3x and 3y=2x