The base `B C` of a ` A B C` is bisected at the point `(p ,q)` & the equation to the side `A B&A C` are `p x+q y=1` & `q x+p y=1` . The equation of the median through `A` is: `(p-2q)x+(q-2p)y+1=0` `(p+q)(x+y)-2=0` `(2p q-1)(p x+q y-1)=(p^2+q^2-1)(q x+p y-1)` none of these

{:(2x + 3y = 9),((p + q)x + (2p - q)y = 3(p + q+ 1)):}

The point of injtersection of the line x/p+y/q=1 and x/q+y/p=1 lies on the line

The lines (p-q)x+(q-r)y+(r-p)=0(q-r)x+(r-p)y+(p-q)=0,(x-p)x+(p-q)y+(q-r)=

If (x_1, y_1)&(x_2, y_2) are the ends of diameter of a circle such that x_1&x_2 are the roots of the equation a x^2+b x+c=0 and y_1&y_2 are the roots of the equation p y^2+q y+c=0 . Then the coordinates f the centre of the circle is: (b/(2a), q/(2p)) (b) (-b/(2a),=q/(2p)) (b/a , q/p) (d) None of these

If (x_(1),y_(1))&(x_(2),y_(2)) are the ends of a diameter of a circle such that x_(1)&x_(2) are the roots of the equation ax^(2)+bx+c=0 and y_(1)&y_(2) are the roots of the equation py^(2)=qy+c=0. Then the coordinates of the centre of the circle is: ((b)/(2a),(q)/(2p))(-(b)/(2a),(q)/(2p))((b)/(a),(q)/(p)) d.none of these

Equation of any circle passing through the point(s) of intersection of circle S=0 and line L=0 is S + kL = 0 . Let P(x_1, y_1) be a point outside the circle x^2 + y^2 = a^2 and PA and PB be two tangents drawn to this circle from P touching the circle at A and B . On the basis of the above information : The circle which has for its diameter the chord cut off on the line px+qy - 1 = 0 by the circle x^2 + y^2 = a^2 has centre (A) (p/(p^2 + q^2), (-q)/(p^2 + q^2) (B) (p/(p^2 + q^2), (q)/(p^2 + q^2) (C) (p/(p^2 + q^2), (q)/(p^2 + q^2) (D) none of these

The roots of the equation (q-r)x^2+(r-p)x+p-q=0 are (A) (r-p)/(q-r),1 (B) (p-q)/(q-r),1 (C) (q-r)/(p-q),1 (D) (r-p)/(p-q),1