Angle Between Lines and Parallel/Perpendicular Lines

Hello JEE aspirants! This lesson is crucial for mastering coordinate geometry. You'll learn how to find the angle between two lines and the conditions for them to be parallel or perpendicular. These concepts are fundamental and appear frequently in JEE Main, both directly and indirectly within other problems. So, let's dive in!

1. Angle Between Two Lines

Imagine two lines intersecting on a coordinate plane. They form two pairs of vertically opposite angles. Our focus is on finding one of these angles (usually the acute one) using the slopes of the lines.

Let's say we have two lines with slopes $m_1$ and $m_2$ . The slope represents the tangent of the angle each line makes with the positive x-axis. Geometrically, the angle between the lines is related to the difference in these angles.

The formula to find the angle $\theta$ between the lines is:

\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|

Derivation: Let $\theta_1$ and $\theta_2$ be the angles made by the lines with slopes $m_1$ and $m_2$ respectively, with the positive x-axis. Then, $m_1 = \tan\theta_1$ and $m_2 = \tan\theta_2$ . The angle between the lines, $\theta = |\theta_1 - \theta_2|$ . Thus, $\tan\theta = |\tan(\theta_1 - \theta_2)| = \left|\frac{\tan\theta_1 - \tan\theta_2}{1 + \tan\theta_1 \tan\theta_2}\right| = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$ . The absolute value ensures that we get the acute angle.

Example: Consider two lines with slopes $m_1 = 2$ and $m_2 = -\frac{1}{3}$ . The angle between them is $\tan\theta = \left|\frac{2 - (-\frac{1}{3})}{1 + 2(-\frac{1}{3})}\right| = \left|\frac{\frac{7}{3}}{\frac{1}{3}}\right| = 7$ . Therefore, $\theta = \tan^{-1}(7)$ .

Tip: Always take the absolute value. If $\tan\theta$ is negative, you're finding the obtuse angle. The acute angle will be $180^\circ$ minus the obtuse angle. However, using the absolute value directly gives you the acute angle!

2. Condition for Parallel Lines

Two lines are parallel if they never intersect. Geometrically, this means they have the same inclination with the x-axis. Algebraically, this translates to their slopes being equal.

m_1 = m_2 \text{ (parallel lines)}

Explanation: If the slopes are equal, the tangent of the angles they make with the x-axis is the same. Hence, the angles themselves are equal, ensuring the lines are parallel. Consider the lines $y = 2x + 3$ and $y = 2x - 1$ . They both have a slope of $2$ , and therefore are parallel.

3. Condition for Perpendicular Lines

Two lines are perpendicular if they intersect at a right angle ( $90^\circ$ ). The slopes of perpendicular lines have a special relationship: the product of their slopes is -1.

m_1 \times m_2 = -1 \text{ (perpendicular lines)}

Explanation: If $m_1 \times m_2 = -1$ , then $m_2 = -\frac{1}{m_1}$ . Geometrically, this means that if one line has a slope $m$ , the other line has a slope that is the negative reciprocal of $m$ . For instance, if a line has a slope of $3$ , a line perpendicular to it will have a slope of $-\frac{1}{3}$ . Consider the lines $y = 2x + 5$ and $y = -\frac{1}{2}x + 1$ . Since $2 \times (-\frac{1}{2}) = -1$ , these lines are perpendicular.

Derivation: If $\theta = 90^{\circ}$ , then $\tan\theta$ is undefined. This happens when the denominator of the $\tan\theta$ formula is zero: $1 + m_1 m_2 = 0$ . Therefore, $m_1 m_2 = -1$ .

4. Equation of a Line Parallel to a Given Line

If you have a line $ax + by + c = 0$ , any line parallel to it will have the same coefficients for $x$ and $y$ but a different constant term. This is because the slope, $-\frac{a}{b}$ , remains the same.

\text{Line parallel to } ax + by + c = 0: ax + by + k = 0

Explanation: Changing the constant term shifts the line up or down, preserving its slope. For example, given the line $2x + 3y + 4 = 0$ , a parallel line will have the form $2x + 3y + k = 0$ . The value of $k$ determines the specific position of the parallel line.

5. Equation of a Line Perpendicular to a Given Line

For a line $ax + by + c = 0$ , a perpendicular line is obtained by swapping the coefficients of $x$ and $y$ , changing the sign of one of them, and using a different constant term. This ensures the product of the slopes is -1.

\text{Line perpendicular to } ax + by + c = 0: bx - ay + k = 0

Explanation: The slope of the original line is $-\frac{a}{b}$ . The slope of the new line is $\frac{b}{a}$ . Their product is $-\frac{a}{b} \times \frac{b}{a} = -1$ . For example, given the line $2x + 3y + 4 = 0$ , a perpendicular line will have the form $3x - 2y + k = 0$ .

Tip: To find the specific values of $k$ in the parallel and perpendicular line equations, you'll usually be given an additional point that the line must pass through. Substitute the coordinates of that point into the equation and solve for $k$ .

Common Mistake: Forgetting the absolute value in the angle between lines formula! This leads to finding the obtuse angle instead of the acute angle. Also, be careful with signs when determining perpendicular slopes.

JEE Specific Trick: Many JEE problems involve finding the equation of a line that's parallel or perpendicular to another line and also satisfies some other geometric condition (e.g., passes through a given point, is tangent to a circle, etc.). Break the problem down into smaller steps:

Write down the general equation of the parallel/perpendicular line using $k$ .
Use the other geometric condition to find the value of $k$ .
Substitute the value of $k$ back into the equation.

Example: Find the equation of the line that is perpendicular to $2x+3y+4=0$ and passes through the point $(1,1)$ . The perpendicular line is of the form $3x - 2y + k = 0$ . Plugging in $(1,1)$ , we get $3(1)-2(1)+k=0$ , so $k=-1$ . Therefore the equation is $3x - 2y -1 = 0$ .

That wraps up our lesson on angles between lines and parallel/perpendicular lines. Remember to practice lots of problems to solidify these concepts. Good luck with your JEE preparation!