If the equation 2 x^2+k x y-6 y^2+3 x+y+1=0 , (k>0) represents a pair of straight lines, then their point of intersection is

Step-by-step solution

Given equation, 2 x^2+k x y-6 y^2+3 x+y+1=0 is a pair of straight line. | array ccc 2 & k / 2 & 3 / 2 k / 2 & -6 & 1 / 2 3 / 2 & 1 / 2 & 1 array |=0 | array ccc 4 & k & 3 k & -12 & 1 3 & 1 & 2 array |=0=4(-24-1)-k(2 k-3)+3(k+36)=0 -100-2 k^2+3 k+3 k+108=0 array ll & 2 k^2-6 k-8=0 & k^2-3 k-4=0 array (k-4)(k+1)=0, k=4 k>0 Equation of line is 2 x^2+4 x y-6 y^2+3 x+y+1=0 array rrr & (2 x-2 y+1)(x+3 y+1) & =0 & 2 x-2 y+1=0 and x+3 y+1 & =0 array Solving equation, we get x= -5 8 , y= -1 8 Intersection point ( -5 8 , -1