Q.Show that $2x^2+5xy+2y^2=0$ represents a pair of straight lines, find the separate equations of the two lines, and find the angle between them.

This is a homogeneous second-degree equation, so it automatically represents a pair of lines through the origin; factoring it directly gives the two individual lines, and their angle can then be checked two ways.

Since every term of $2x^2+5xy+2y^2=0$ has degree exactly $2$, it is homogeneous, and therefore automatically represents a pair of straight lines through the origin (comparing with $ax^2+2hxy+by^2=0$: $a=2$, $2h=5$ so $h=\dfrac52$, $b=2$).

Factoring. We look for factors $(l_1x+m_1y)(l_2x+m_2y)$ that expand to $2x^2+5xy+2y^2$. Trying $(2x+y)(x+2y)$: $2x \cdot x = 2x^2$; $2x\cdot 2y=4xy$; $y \cdot x = xy$; $y \cdot 2y = 2y^2$; total $= 2x^2+5xy+2y^2$ ✓. So the two lines are:

Angle (via the two separate lines). Slope of $2x+y=0$ is $m_1=-2$. Slope of $x+2y=0$ is $m_2=-\dfrac12$.

$\tan\theta = \left|\dfrac{m_1-m_2}{1+m_1m_2}\right| = \left|\dfrac{-2-(-\frac12)}{1+(-2)(-\frac12)}\right| = \left|\dfrac{-\frac32}{2}\right| = \dfrac34$ …

Direct formula method: computing $\tan\theta$ from $a,h,b$ without factoring first (shown as the independent check above) is itself a valid, faster alternative …

A common factoring error is trying $(2x+2y)(x+y)$ or similar mismatched pairs that do not reproduce the correct middle term $5xy$; always expand a proposed factoring fully to …