Bisector Containing the Origin

If a₁·a₂ + b₁·b₂ > 0, this bisector (+ sign, with c₁, c₂ same sign) contains the origin. Reverse if < 0.

The two bisectors of $L_1: a_1x+b_1y+c_1=0$ and $L_2: a_2x+b_2y+c_2=0$ are:

$$\frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}} = +\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}} \qquad \text{and} \qquad \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}} = -\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$$

To determine which bisector contains the origin, substitute $(0,0)$:

$$S_1 = \frac{c_1}{\sqrt{a_1^2+b_1^2}}, \qquad S_2 = \frac{c_2}{\sqrt{a_2^2+b_2^2}}$$

• If $S_1$ and $S_2$ have the same sign: the origin satisfies $S_1 = +S_2$, so it lies on the $+$ bisector.
• If $S_1$ and $S_2$ have opposite signs: the origin lies on the $-$ bisector.

For an arbitrary point $Q(h,k)$

Compute $\dfrac{a_1h+b_1k+c_1}{\sqrt{a_1^2+b_1^2}}$ and $\dfrac{a_2h+b_2k+c_2}{\sqrt{a_2^2+b_2^2}}$.

Same rule: same sign $\Rightarrow$ $Q$ is on the $+$ bisector; opposite signs $\Rightarrow$ $Q$ is on the $-$ bisector.

This is the standard JEE test: identify the bisector containing a specific point by a single sign check, without solving the bisector equations.