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

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: (a) (p-2q)x+(q-2p)y+1=0 (b) (p+q)(x+y)-2=0 (c) (2p q-1)(p x+q y-1)=(p^2+q^2-1)(q x+p y-1) (d)none of these

The base BC of a hat ABC is bisected at the point (p,q)& the equation to the side AB&AC are px+qy=1&qx+py=1. The equation of the median through A is: (p-2q)x+(q-2p)y+1=0(p+q)(x+y)-2=0(2pq-1)(px+qy-1)=(p^(2)+q^(2)-1)(qx+py-1) none of these

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

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

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 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 of the centre of the circle is: (b/(2a), q/(2p)) ( b/(2a), q/(2p)) (b/a , q/p) d. none of these

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 the points (p,0), (0,q)and (x,y) are collinear, prove that x/p+y/q= 1 .

If two distinct chords, drawn from the point (p, q) on the circle x^2+y^2=p x+q y (where p q!=q) are bisected by the x-axis, then p^2=q^2 (b) p^2=8q^2 p^2 8q^2

If two distinct chords, drawn from the point (p, q) on the circle x^2+y^2=p x+q y (where p q!=q) are bisected by the x-axis, then (a) p^2=q^2 (b) p^2=8q^2 p^2 8q^2