1. Let PQR be a right angled isosceles triangle, right angled at P (2,1). If the equation of the line QR is 2x + y 3, then the equation representing the pair of lines P and PR is

Options

A. `3x^(2)-3y^(2)+8xy+20x+10y+25=0`
B. `3x^(2)-3y^(2)+8xy-20x-10y+25=0`
C. `3x^(2)-3y^(2)+8xy+10x+15y+20=0`
D. `3x^(2)-3y^(2)-8xy-10x-15y-20=0`

Explanation

Let `S` be the mid-point of `QR` and given `DeltaPQR` is an isosceles.
Therefore, `PSbotQR` and `S` is mid-point of hypotenuse, therefore `S` is equidistant from `P`, `Q`, `R`.
`:. PS=QS=RS`
Since, `/_P=90^(@)` and `/_Q=/_R`
But `/_P+/_Q+/_R=180^(@)`
`:.90^(@)+/_Q+/_R=180^(@)`
`implies /_Q=/_R=45^(@)`

Now, slope of `QR` is `-2` [given]
But `QRbotPS`.
`:.` Slope of `PS` is `1//2`.
Let `m` be the slope of `PQ`.
`:.tan(+-45^(@))=(m-1//2)/(1-m(-1//2))`
`implies+-1=(2m-1)/(2+m)`
`impliesm=3,-1//3`
`:.` Equations of `PQ` and `PR` are
`y-1=3(x-2)`
and `y-1=-(1)/(3)(x-2)`
or `3(y-1)+(x-2)=0`
Therefore, joint equation of `PQ` and `PR` is
`[3(x-2)-(y-1)][(x-2)+3(y-1)]=0`
`implies3(x-2)^(2)-3(y-1)^(2)+8(x-2)(y-1)=0`
`implies3x^(2)-3y^(2)+8xy-20x-10y+25=0`