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 PQ and PR is
3x2−3y2+8xy−20x−10y+25=0
Let the slopes of PQ and PR be m and −1m respectively. Since PQR is an isosceles triangle ∠PQR=∠PRQ
⇒ ∣∣m+21−2m∣∣=∣∣∣−1m+21+2m∣∣∣
[∵ slope of QR = -2]
⇒ m+2=±(1−2m)⇒m=3 or −13
So the equations of PQ and PR are
(y−1) = 3(x−2) and y−1=(−13)(x−2)