Angle Bisectors and Family of Lines

Hello JEE aspirants! This lesson dives into angle bisectors, family of lines, and concurrency – crucial concepts for tackling coordinate geometry problems in JEE Main. Mastering these will significantly boost your problem-solving speed and accuracy.

1. Equation of Angle Bisectors of Two Lines

Consider two lines represented by the equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ . An angle bisector is a line that divides the angle between these two lines into two equal halves. Geometrically, any point on the angle bisector is equidistant from both lines.

The equation of the angle bisectors is given by the formula:

\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}

Explanation: The formula arises directly from the condition that the perpendicular distance from a point $(x, y)$ on the bisector to both lines must be equal. The $\pm$ sign gives us two bisectors, which are perpendicular to each other.

Example: Let's say you have lines $3x + 4y + 5 = 0$ and $12x - 5y - 2 = 0$ . You'd plug the coefficients into the formula above and solve for the two equations, one with "+" and the other with "-". Each of these is an angle bisector.

2. Bisector of Acute and Obtuse Angles

The two angle bisectors we found are perpendicular, but how do we determine which bisector corresponds to the acute angle and which to the obtuse angle?

First, make sure that $c_1$ and $c_2$ are positive. If not, multiply the equation by $-1$ .
Determine the sign of $a_1a_2 + b_1b_2$ $a_{1} a_{2} + b_{1} b_{2}$ .
- If $a_1a_2 + b_1b_2 > 0$ , then the bisector with the "+" sign is the bisector of the obtuse angle, and the bisector with the "-" sign is the bisector of the acute angle.
- If $a_1a_2 + b_1b_2 < 0$ , then the bisector with the "+" sign is the bisector of the acute angle, and the bisector with the "-" sign is the bisector of the obtuse angle.

Geometric Intuition: The sign of $a_1a_2 + b_1b_2$ essentially tells us about the angle between the normal vectors of the two lines. If the dot product is positive, the angle is acute; if negative, it's obtuse.

3. Bisector Containing the Origin

Sometimes, we need to find the bisector that contains the origin. This is particularly important when dealing with triangles or regions defined around the origin.

To find the bisector containing the origin, ensure that $c_1 > 0$ and $c_2 > 0$ . Then, choose the bisector for which the constant term is positive.

Why? This ensures that the region containing the origin satisfies the inequality defined by the line. Consider $ax + by + c > 0$ . If $(0, 0)$ lies in this region, $c > 0$ .

4. Family of Lines Through the Intersection Point

If we have two lines, $L_1: a_1x + b_1y + c_1 = 0$ and $L_2: a_2x + b_2y + c_2 = 0$ , then any line passing through their intersection point can be represented as a linear combination of $L_1$ and $L_2$ . This is called the "family of lines."

\text{Family: } (a_1x + b_1y + c_1) + \lambda(a_2x + b_2y + c_2) = 0

Explanation: The parameter $\lambda$ can take any real value, generating different lines passing through the point of intersection of $L_1$ and $L_2$ . Geometrically, as $\lambda$ varies, it "rotates" the line around the intersection point.

Example: If you're given that a line passes through the intersection of $x + y - 1 = 0$ and $2x - y + 3 = 0$ and also has a slope of 2, you can set up the equation $(x + y - 1) + \lambda(2x - y + 3) = 0$ . Then, rearrange into $y = mx + c$ form, set $m = 2$ , and solve for $\lambda$ .

5. Concurrency of Three Lines

Three or more lines are said to be concurrent if they pass through the same point. To check if three lines, $a_1x + b_1y + c_1 = 0$ , $a_2x + b_2y + c_2 = 0$ , and $a_3x + b_3y + c_3 = 0$ , are concurrent, we use the determinant condition:

\text{Concurrency: } \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0

Explanation: This determinant being zero is equivalent to saying that the system of equations has a non-trivial solution, meaning there exists a common point satisfying all three equations. In other words, the three lines intersect at a single point.

Example: If you need to prove three lines are concurrent, calculate this determinant using the coefficients of their equations. If the determinant is zero, the lines are concurrent.

Tip: When dealing with angle bisectors, remember to normalize the coefficients (divide by

\sqrt{a^2 + b^2}

). This ensures you're working with actual distances, not just expressions.

Common Mistake: Forgetting to consider both the positive and negative signs when finding angle bisectors. Always remember there are two bisectors!

JEE Trick: Often, JEE problems involve a combination of these concepts. For instance, finding the equation of a line that belongs to a family of lines and is also an angle bisector. Practice identifying these "combined concept" problems. When you see a problem that combines multiple elements, break it down into the individual concepts you've learned.

Good luck, and keep practicing! You've got this!