[x-2y=0],[3x+4y=10]

Prove that the three lines 5x+3y-7=0, 3x-4y=10, and x+2y=0 meet in a point.

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

The area of the parallelogram formed by lines 2x-y+3=0, 3x+4y-6=0,2x-y+9=0, 3x+4y+4=0 is (in sq units)

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

2x+3y=0 3x+4y=5

The perimeter of a parallelogram whose sides are represented by the lines x+2y+3=0 , 3x+4y-5=0,2x+5=0 and 3x+4y-10=0 is equal to

The value of 'lambda' so that the system of equations x+2y+3=0 3x+4y+4=0 7x-10 y+lambda=0 is consistent, is

The perimeter of a parallelogram whose sides are represented by the lines x+2y+3=0 , 3x+4y-5=0,2x+4y+5=0 and 3x+4y-10=0 is equal to

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