[2x+y-6=0],[4x-2y-4=0]

The two lines 2x+y-6=0 and 4x-2y-4=0 intersect at……

Check whether the following equations are consistent or inconsistent. Solve them graphically. 2x+y-6=0 and 4x-2y-4=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 equations x+2y-3=0, 3x+4y-7=0, 2x+3y-4=0 and 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 x+2y-5=0, 2x-3y+4=0,6x+4y-13=0

Equations of the common tangents to the parabola, y=x^(2) and y=-(x-2)^(2) are [x=0,x=4y-2],[x=0,y=4x-2],[y=0,y=4x+4],[y=0,y=4x-4]

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 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)