[px+2y=5],[3x+y=1]

The pair of linear equations px+2y=5 and 3x+y=1 has unique solution if

{:(kx + 2y = 5),(3x + y = 1):}

3x-y=2,5x-2y=1

Find the value of p if the lines 3x+4y=5, 2x+3y=4, px+4y=6 are concurrent.

Prove that [[x, x^2 , 1+px^3], [y, y^2, 1+py^3] ,[z, z^2, 1+pz^3]] = (1+pxyz)(x-y)(y-z)(z-x)

If the lines 2x+3y=8,5x-6y+7=0 and px+py=1 are concurrent,then the line x+2y-1=0 passes through

Solve for x and y 2/(3x+2y)+3/(3x-2y)=17/5,5/(3x+2y)+1/(3x+2y)=6/5

Find the value of p, so that three lines 3x + y = 2, px + 2y-3=0 and 2x-y=3 are concurrent.

Let the straight lines x+y-2=0, 2x-y+1=0 and px+qy-r=0 be concurrent and l_1 and l_2 be the two members of the family of lines 2px+2qy+4r=0 which are nearest and farthest from origin : Now answer the following questions : The eqation of line l_1 is : (A) y=5x (B) y=3x (C) 5y=x (D) none of these