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The base B C of a A B C is bisected at ...

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

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)`

Text Solution

Verified by Experts

The correct Answer is:
C
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)`
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