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

Options

A. `qx - py = 0`
B. `(x)/(p) +(y)/(q) = 2`
C. `(2pq -1) (px +qy-1) =(p^(2)+q^(2)-1) (qx+py-1)`
D. `(p-2q)x+(q-2p)y = p^(2) +r^(2)`

Answer

C

Explanation

Equation of the line through Ab and AC is
`(px +qy -1) +lambda (qx +py -1) =0` (i)
`:'` It passes through (p,q), then
`(p^(2)+q^(2)-1) +lambda (pq +pq -1) =0`
`:. lambda =- ((p^(2)+q^(2)-1))/((2pq -1))`
`:.` From equation (i),
`(px +qy -1) -((p^(2)+q^(2)-1))/((2pq-1)) (qx +py -1) =0`
or `(2pq -1) (px +qy -1) =(p^(2) +q^(2)-1) (qx+py -1)`