Q.Show that $2x^2 + 3xy - 2y^2 + 3x + y + 1 = 0$ represents a pair of perpendicular lines.

The angle between the lines of $ax^2+2hxy+by^2+\ldots=0$ satisfies $\tan\theta=\left|\dfrac{2\sqrt{h^2-ab}}{a+b}\right|$, so the lines are perpendicular exactly when $a+b=0$, independent of $h$.

Step 1. Identify $a,b$. Comparing $2x^2+3xy-2y^2+3x+y+1=0$ with the general form: $a=2,\ 2h=3\Rightarrow h=\tfrac32,\ b=-2$.

Step 2. Test perpendicularity.

so if this equation represents a genuine pair of lines, they must be perpendicular.

Step 3. Factor the quadratic part.

since $(2x-y)(x+2y)=2x^2+4xy-xy-2y^2=2x^2+3xy-2y^2$. ✓

Step 4. Find the constants by comparing coefficients. Write $(2x-y+p)(x+2y+q)=2x^2+3xy-2y^2+(2q+p)x+(p-3q)\,y$...

more directly: expand $(2x-y+p)(x+2y+q)=2x^2+4xy+2qx-xy-2y^2-qy+px+2py+pq$

Wait — carefully collecting the $y$-coefficient: $-q+2p=(2p-q)$. So matching against $3x+y+1$:

From the first, $p=3-2q$. Substitute into the second: $2(3-2q)-q=1\Rightarrow 6-5q=1\Rightarrow q=1$, hence $p=1$. Check: $pq=1\times1=1$ ✓, matching the constant term.

Step 5. Write the factors.

This is a genuine product of two distinct lines, confirming the equation does represent a real pair.

Step 6. Verify perpendicularity directly. Line $2x-y+1=0$ has slope $2$; line $x+2y+1=0$ has slope $-\tfrac12$. Product of slopes $=2\times\left(-\tfrac12\right)=-1$, confirming perpendicularity.

• Concluding perpendicularity from $a+b=0$ alone without checking the equation is a genuine (factorisable) pair of lines
• Sign slip when comparing the linear coefficients $2q+p$ and $2p-q$