Question

If coordinate axes are the angle bisectors of the pair of lines

Answer 1

Let's consider an easier case, when the given lines have the form y=mxy=mx and y=nxy=nx (with m≠nm≠n, of course).

The points on the angle bisectors satisfy the conditions that their distance from the two lines is the same. So you have, for such a point (x,y)(x,y),

|y−mx|1+m2−−−−−−√=|y−nx|1+n2−−−−−√|y−mx|1+m2=|y−nx|1+n2

By squaring, we get

(y−mx)2(1+n2)=(y−nx)2(1+m2)(y−mx)2(1+n2)=(y−nx)2(1+m2)

and expanding

(y−mx)2−(y−nx)2=m2(y−nx)2−n2(y−mx)2.(y−mx)2−(y−nx)2=m2(y−nx)2−n2(y−mx)2.

The left hand side becomes

(y−mx+y−nx)(y−mx−y+nx)=−(m−n)(2y−(m+n)x)x(y−mx+y−nx)(y−mx−y+nx)=−(m−n)(2y−(m+n)x)x

and the right hand side is

(my−mnx+ny−mnx)(my−mnx−ny+mnx)=(m−n)((m+n)y−2mnx)y(my−mnx+ny−mnx)(my−mnx−ny+mnx)=(m−n)((m+n)y−2mnx)y

Since m≠nm≠n, we can factor out m−nm−non both sides, getting

−2xy+(m+n)x2=(m+n)y2−2mnxy−2xy+(m+n)x2=(m+n)y2−2mnxy

that can be written

(m+n)x2−2(1−mn)xy−(m+n)y2=0(m+n)x2−2(1−mn)xy−(m+n)y2=0

If you consider the equation by2+2hxy+ax2=0by2+2hxy+ax2=0 in the unknown yy, you can see that mm and nn are the roots of it (when b≠0b≠0). Thus m+n=−2h/bm+n=−2h/b and mn=a/bmn=a/b, so you can rewrite the above equation in the form

−2hbx2−2(1−ab)xy−−2hby2=0−2hbx2−2(1−ab)xy−−2hby2=0

which is

hx2−(a−b)xy−hy2=0hx2−(a−b)xy−hy2=0

Note that it can't be both h=0h=0 and a=b≠0a=b≠0, because in this case the original equations would be x2+y2=0x2+y2=0(the isotropic lines, if you consider the complex plane).