Basics and Forms of Straight Line Equation

Basics and Forms of Straight Line Equation

Hey there, JEE aspirants! Welcome to the world of straight lines. Coordinate geometry is super important for JEE Main, and mastering straight lines is the foundation. It's not just about memorizing formulas; it's about understanding the geometry and applying the concepts. Let's dive in and make sure you're ready to tackle any straight line problem!

1. Slope of a Line

The slope of a line, often denoted by $m$, tells you how steep the line is. Geometrically, it's the tangent of the angle the line makes with the positive x-axis, measured counter-clockwise.

Formula:

Here, $(x_1, y_1)$ and $(x_2, y_2)$ are any two points on the line, and $\theta$ is the angle the line makes with the positive x-axis.

Intuition: Think of slope as "rise over run." If you move from point $(x_1, y_1)$ to $(x_2, y_2)$, the 'rise' is the change in $y$ values ($y_2 - y_1$), and the 'run' is the change in $x$ values ($x_2 - x_1$).

Example: If a line passes through points (1, 2) and (3, 6), the slope is $$m = \frac{6 - 2}{3 - 1} = \frac{4}{2} = 2$$. This means for every 1 unit you move along the x-axis, you move 2 units along the y-axis.

2. Slope-Intercept Form

The slope-intercept form is one of the most common ways to represent a straight line. It directly tells you the slope and where the line intersects the y-axis.

Formula:

Here, $m$ is the slope and $c$ is the y-intercept (the y-coordinate of the point where the line crosses the y-axis).

Derivation: Consider a line with slope $m$ that intersects the y-axis at $(0, c)$. Take any other point $(x, y)$ on the line. Using the slope formula, we have $$m = \frac{y - c}{x - 0}$$, which simplifies to $y = mx + c$.

Example: The equation $y = 3x + 5$ represents a line with a slope of 3 and a y-intercept of 5. So, the line passes through the point (0, 5).

3. Point-Slope Form

If you know the slope of a line and a point it passes through, you can use the point-slope form to find the equation of the line.

Formula:

Here, $m$ is the slope, and $(x_1, y_1)$ is a known point on the line.

Derivation: Similar to the slope-intercept form, consider a line with slope $m$ passing through the point $(x_1, y_1)$. Take any other point $(x, y)$ on the line. Using the slope formula, we have $$m = \frac{y - y_1}{x - x_1}$$, which rearranges to $y - y_1 = m(x - x_1)$.

Example: Find the equation of a line that passes through the point (2, -3) and has a slope of -1. Using the point-slope form: $$y - (-3) = -1(x - 2)$$, which simplifies to $y + 3 = -x + 2$, or $y = -x - 1$.

4. Two-Point Form

If you know two points on a line, you can find its equation using the two-point form. This form is derived by first finding the slope using the two points and then applying the point-slope form.

Formula:

Here, $(x_1, y_1)$ and $(x_2, y_2)$ are the two known points on the line.

Derivation: The slope of the line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $$m = \frac{y_2 - y_1}{x_2 - x_1}$$. Now, using the point-slope form with the point $(x_1, y_1)$, we get $$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$, which rearranges to the two-point form.

Example: Find the equation of a line passing through (1, 4) and (3, 8). Using the two-point form: $$\frac{y - 4}{8 - 4} = \frac{x - 1}{3 - 1}$$, which simplifies to $$\frac{y - 4}{4} = \frac{x - 1}{2}$$. Cross-multiplying gives $$2(y - 4) = 4(x - 1)$$, or $$2y - 8 = 4x - 4$$, which simplifies to $$y = 2x + 2$$.

5. Intercept Form

The intercept form focuses on where the line intersects the x and y axes.

Formula:

Here, $a$ is the x-intercept (the x-coordinate of the point where the line crosses the x-axis), and $b$ is the y-intercept.

Derivation: The line passes through $(a, 0)$ and $(0, b)$. Using the two-point form: $$\frac{y - 0}{b - 0} = \frac{x - a}{0 - a}$$, which simplifies to $$\frac{y}{b} = \frac{x - a}{-a}$$. Rearranging this gives $$\frac{x}{a} + \frac{y}{b} = 1$$.

Example: If a line has an x-intercept of 2 and a y-intercept of 3, its equation is $$\frac{x}{2} + \frac{y}{3} = 1$$.

6. Normal Form

The normal form uses the perpendicular distance from the origin to the line and the angle this perpendicular makes with the x-axis.

Formula:

Here, $p$ is the perpendicular distance from the origin to the line, and $\alpha$ is the angle that the perpendicular makes with the positive x-axis.

Explanation: This form is a bit trickier to derive directly. Imagine a line, and draw a perpendicular from the origin to that line. This perpendicular has length $p$ and makes an angle $\alpha$ with the x-axis. Any point $(x, y)$ on the line must satisfy the equation $x\cos\alpha + y\sin\alpha = p$.

Example: If the perpendicular distance from the origin to a line is 4, and the angle the perpendicular makes with the x-axis is 30 degrees, the equation is $$x\cos(30^\circ) + y\sin(30^\circ) = 4$$, which is $$x\frac{\sqrt{3}}{2} + y\frac{1}{2} = 4$$.

General Form

The general form of a straight line equation is:

Formula:

Where $a$, $b$, and $c$ are constants. All other forms can be converted to this form. To convert from general form to normal form divide the entire equation by $\sqrt{a^2 + b^2}$. So dividing by $\sqrt{a^2 + b^2}$ is also a step towards converting it into normal form.

That's it for the basics and forms of straight line equations. Practice lots of problems, visualize the geometry, and you'll ace this topic in JEE Main!