Consider the equation, x2+2√2xy+ky2 Then tanθ=
2√2−k
k+1
tan45∘=
2√2−k
k+1
(k+1)2=4(2−k) k2+2k+1=8−4k Further simplify the above, k2+6k−7=0 (k+7)(k−1)=0 k=1,k≠−7 Now the equation of the angle bisector of the line is given by, x2+2√2xy+y2=0
x2−y2
0
=
xy
√2
x2−y2=0 The angle of the triangle formed by the lines x2−y2=0 and x+2y+1=0 is given by ∆=
