Let the equation of the pair of lines, $$y = px$$ and $$y = qx$$, can be written as $$(y - px)(y - qx) = 0$$. Then the equation of the pair of the angle bisectors of the lines $$x^2 - 4xy - 5y^2 = 0$$ is:

We start with the given pair of lines

$$x^{2}-4xy-5y^{2}=0.$$

Because the constant term is zero, the curve passes through the origin, so it must represent two straight lines through the origin. We factor the quadratic:

$$x^{2}-4xy-5y^{2}= (x-5y)(x+y)=0.$$

Hence the two individual lines are

$$L_{1}:x-5y=0 \quad\Longrightarrow\quad y=\frac{1}{5}x,$$

$$L_{2}:x+y=0 \quad\Longrightarrow\quad y=-x.$$

Now we wish to obtain the equations of the angle-bisector lines of these two lines. For two lines through the origin,

$$a_{1}x+b_{1}y=0 \quad\text{and}\quad a_{2}x+b_{2}y=0,$$

the angle-bisectors are obtained from the formula

$$\frac{a_{1}x+b_{1}y}{\sqrt{a_{1}^{2}+b_{1}^{2}}}= \pm \frac{a_{2}x+b_{2}y}{\sqrt{a_{2}^{2}+b_{2}^{2}}}.$$

Here we have

$$L_{1}:a_{1}=1,\;b_{1}=-5, \qquad L_{2}:a_{2}=1,\;b_{2}=1.$$

Therefore

$$\frac{x-5y}{\sqrt{1^{2}+(-5)^{2}}}= \pm \frac{x+y}{\sqrt{1^{2}+1^{2}}}.$$

Simplifying the square-roots,

$$\frac{x-5y}{\sqrt{26}} = \pm \frac{x+y}{\sqrt{2}}.$$

Cross-multiplying gives

$$\sqrt{2}\,(x-5y)= \pm \sqrt{26}\,(x+y).$$

To obtain a single equation containing both bisectors, we square both sides and then bring all terms to one side:

$$\bigl[\sqrt{2}\,(x-5y)\bigr]^{2} - \bigl[\sqrt{26}\,(x+y)\bigr]^{2}=0.$$

That is,

$$2\,(x-5y)^{2} - 26\,(x+y)^{2}=0.$$

Dividing every term by $$2$$ for convenience,

$$(x-5y)^{2} - 13\,(x+y)^{2}=0.$$

Now we expand each square separately:

First square: $$ (x-5y)^{2}=x^{2}-10xy+25y^{2}. $$

Second square: $$ (x+y)^{2}=x^{2}+2xy+y^{2}. $$

Substituting these into the equation, we get

$$x^{2}-10xy+25y^{2}-13\bigl(x^{2}+2xy+y^{2}\bigr)=0.$$

Distribute the $$-13$$:

$$x^{2}-10xy+25y^{2}-13x^{2}-26xy-13y^{2}=0.$$

Combine like terms for each power:

For $$x^{2}: \;\; x^{2}-13x^{2}= -12x^{2},$$

For $$xy:\;\; -10xy-26xy= -36xy,$$

For $$y^{2}:\;\; 25y^{2}-13y^{2}= 12y^{2}.$$

Thus the equation becomes

$$-12x^{2}-36xy+12y^{2}=0.$$

Every coefficient is divisible by $$-12$$. Dividing through by $$-12$$ gives the simplest form:

$$x^{2}+3xy-y^{2}=0.$$

This is exactly the required combined equation of the two angle-bisector lines. Matching this with the options, we see it corresponds to Option C.

Hence, the correct answer is Option C.