Q.If the equation $S \equiv ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of parallel straight lines, then show that

Once the determinant test of the previous concept confirms that $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a genuine pair of straight lines, two natural follow-up questions arise depending on whether the lines are parallel or intersecting.

If the lines are parallel ($h^2=ab$, so they share the same homogeneous part), we may write them as $lx+my+c_1=0$ and $lx+my+c_2=0$ for the same direction $l,m$ but different constants. Multiplying these out and matching coefficients against the general equation gives $a=l^2$, $b=m^2$, $2g=l(c_1+c_2)$, $c=c_1c_2$. The distance between two parallel lines is the classical $\dfrac{|c_1-c_2|}{\sqrt{l^2+m^2}}$; expressing $|c_1-c_2|$ via $(c_1-c_2)^2=(c_1+c_2)^2-4c_1c_2$ and substituting $c_1+c_2=2g/l$, $c_1c_2=c$, $l^2=a$ gives $|c_1-c_2|=2\sqrt{(g^2-ac)/a}$, and since $l^2+m^2=a+b$, the distance becomes

If the lines intersect ($h^2\neq ab$), their point of intersection is the point where both partial derivatives of the quadratic expression vanish simultaneously — a shortcut that works because differentiating a product of two linear factors with respect to $x$ (or $y$) and setting it to zero isolates exactly the point common to both. This gives the pair of simultaneous linear equations

solved by Cramer's rule as $x=\dfrac{hf-bg}{ab-h^2}$, $y=\dfrac{gh-af}{ab-h^2}$.

As worked checks: $4x^2+12xy+9y^2-10x-15y+4=0$ has $a=4,h=6,b=9$ (indeed $h^2=36=ab$, confirming parallel lines), $g=-5,c=4$, giving distance $d=2\sqrt{(25-16)/(4\times13)}=\dfrac{3\sqrt{13}}{13}$. Separately, $x^2-y^2-x+5y-6=0$ has $a=1,h=0,b=-1,g=-\tfrac12,f=\tfrac52$; solving $x-\tfrac12=0$, $-y+\tfrac52=0$ gives intersection point $\left(\tfrac12,\tfrac52\right)$. …