`y=p(x) and y=q(x)` are lines can be written as `(y-p_x)(y-q_x)=0` then find angle bisector of `x^2-4xy-5y^2=0`

Let the equation of the pair of lines, y = px and y = qx, can be written as (y – px) (y – qx) = 0. Then the equation of the pair of the angle bisectors of the lines x^2 – 4xy – 5y^2 = 0 is:

Find the area of the triangle formed by the line x+y=3 and the angle bisectors of the pair of lines x^(2)-y^(2)+4y-4=0

The equation of the bisectors of the angle between the lines joining the origin to the points of intersection of the curve x^(2)+xy+y^(2)+x+3y+1=0 and the line x+y+2=0 is

If (p)/(x) +(q)/(y) =m and (q)/(x) +(p)/(y) =n , then what is (x)/(y) equal to ?

Find the equation of the bisectors of the angle between the lines represented by 3x^2-5xy+4y^2=0

P is a point on the parabola y^(2)=4x and Q is Is a point on the line 2x+y+4=0. If the line x-y+1=0 is the perpendicular bisector of PQ, then the co-ordinates of P can be:

P is a point on the circle x^(2)+y^(2)=9Q is a point on the line 7x+y+3=0. The perpendicular bisector of PQ is x-y+1=0 Then the coordinates of P are:

Let P be the point (1,0) and Q be a point on the locus y^(2)=8x. The locus of the midpoint of PQ is y^(2)+4x+2=0y^(2)-4x+2=0x^(2)-4y+2=0x^(2)+4y+2=0

If 5x^(2) + 4xy + y^(2) + 2x + 1= 0 then find the value of x, y