2x-3y=4,3y-x=4

(dy)/(dx)=(2x-3y+4)/(3x+4y-5)

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

For each pair of linear equations given below, drawn graphs and then state, whether the lines drawn are parallel or perpendicular to each other. 2x-3y=6 3x+4y=24 x/4+y/3=1

Solve:- 2x+3y=4 3x-y=-5 .

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

(i) 2x+3y=4;3x-y=-5

The area of the parallelogram formed by the lines 4y-3x-a=0, 3y-4x+a=0, 4y-3x-3a=0, 3y-4x+2a=0 is

If the line kx+4y=6 passes through the point of intersection of the two lines 2x+3y=4 and 3x+4y=5, then : k=

Select the correct option from the given alternatives.If the line kx+4y=6 passes through the point of intersection of the two lines 2x+3y=4 and 3x+4y=5 ,then k=

Match the following. {:("Line, parabola"," "," ""Pole"),("I". 2x-3y+4=0,y^(2)=4x," "(a)(4,6)),("II". 2x+3y-4=0,y^(2)=4x," "(b)(2,3)),("III". x-2y+4=0,y^(2)=6x," "(c)(-2,-3)),("IV". 3x+4y-4=0,x^(2)=4y," "(d)(-3//2,-1)):}