# The eqn $ax^2+2hxy+by^2…$

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.?

• you may try to find the slopes of the pair of st. lines...in terms of h,b... – Sujan Dutta Sep 10 '16 at 15:53
• @Sujan, What is the formula to find the slope of pair of straight lines? – pi-π Sep 10 '16 at 15:59

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 $2\sqrt{\frac{g^2-ac}{h^2+a^2}}$