The equation $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ represents a pair of parallel lines. Prove that the equation of the line mid way between the two parallel lines us $hx+by+f=0$
My Attempt:
Let the lines be $lx+my+n_1=0$ and $lx+my+n_2=0$. Then, $$(lx+my+n_1)(lx+my+n_2)=0$$
Comparing the above equation with $ax^2+2hxy+by^2+2gx+2fy+c=0$, we get: $l^2=a, m^2=b, lm=h, l(n_1+n_2)=2g, m(n_1+n_2)=2f$.
Now, what should I do to complete the proof.?
Hint: dist. between two parallel pair of st.lines...$$2\sqrt{\frac{g^2-ac}{h^2+a^2}}$$ remember here $$h^2=ab$$now try to find the equation of the line midway between the lines $L_1=(lx+my+n_1)$ and $L_2=(lx+my+n_2)$. Dist between $L_1$ and $L_2$ is $2\sqrt{\frac{g^2-ac}{h^2+a^2}}$