The point of intersection of lines L1=x−y−7=0 and L2≡x+y−5=0 is (6,−1) Now, for point A(x1,y1) such that PA=3√2, take the equation of line L1 in symmetric form L1:
x−6
cos
π
4
=
y+1
sin
π
4
=±3√2 ⇒x1=6±3 and y1=−1±3 ∵x1,y1≥0, so A(x1,y1) is (9,2) Similarly, for point B(x2,y2) such that PB=√2 take the equation of line L2 in symmetric form L2:
x−6
−cos
π
4
=
y+1
sin
π
4
=±√2 ⇒x2=6+1 and y2=−1±1 ∵x2,y2≥0, so B(x2,y2) is (5,0) ∴ Slope of line joining origin and point A is m1=
2
9
and slope of line joining origin and point B is m2=0