The locus of the orthocenter of the triangle formed by the line (1+p)x-py+p(1+p) = 0, (1+q)x-qy+q(1+q) = 0 and y =0, whete `p ne q,` is

The locus of the orthocentre of the triangle formed by the lines (1+p) x-py + p(1 + p) = 0, (1 + q)x-qy + q(1 +q) = 0 and y = 0, where p!=*q , is (A) a hyperbola (B) a parabola (C) an ellipse (D) a straight line

Find the angle between the straight lines : px-qy+r=0and(p+q)y+(q-p)x+r=0

If the roots of the quadratics x^2 - qx +p = 0 and x^2 - px + q = 0 (p!= q) differ by a constant, show.that p + q + 4 = 0.

Consider three lines as follows. L_1:5x-y+4=0 L_2:3x-y+5=0 L_3: x+y+8=0 If these lines enclose a triangle A B C and the sum of the squares of the tangent to the interior angles can be expressed in the form p/q , where p and q are relatively prime numbers, then the value of p+q is

The product of the sines of the angles of a triangle is p and the product of their cosines is qdot Show that the tangents of the angles are the roots of the equation q x^3-p x^2+(1+q)x-p=0.

The lines px +qy+r=0, qx + ry + p =0,rx + py+q=0, are concurrant then

The coefficients of x^p and x^q in the expansion of (1+x)^(p+q) are

The vertices of a triangle are (p q ,1/(p q)),(q r ,1/(q r)), and (r p ,1/(r p)), where p ,q and r are the roots of the equation y^3-3y^2+6y+1=0 . The coordinates of its centroid are

If the difference of the roots of the equation x^2-px+q=0 is 1, then show that p^2+4q^2= (1+2q)^2 .

Let's solve the following equations and express the roots in form frac(p)(q) (Where q ne 0 and p,q are two integers). 2x + ? = ?.