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

detailed solution

Correct option is B

Let m be the slope of |PQ. Then, tan  450=|m−(−2)1+m(−2)|=|m+21−2m|∴m+2=1−2m or −1+2m=m+2∴m=−13or m=3Hence, the equation of PQ isy−1=113(x−2)or x+3y−5=0 and the eqation  of PR ism1+11−m1=tan α=−1−m21−m2or m1+m2=(tan α−1)2+(tan α+1)2tan2α−1 −2sec2α×cos2αcos 2α∴−2 sec 2α=−2hor cos 2α=1hor 2cos2α−1=1h cosα=1+h2h  and α=h+1h−1y−1=3(x−2) or 3x−y−5=0 Hence, the combined equation of PR and PR is(x+3y−5)(3x−y−5)=0 or 3x2−3y2+8xy−20x−10y+25=0