Straight Lines Ex 10.2: Different Forms of a Line's Equation

Understanding the Different Forms of a Straight Line's Equation

A straight line can be uniquely determined by certain conditions, such as two points it passes through, a point and its slope, or its intercepts with the axes. Each set of conditions leads to a specific form of its equation. Let's explore these essential forms:

1. Point-Slope Form: If a line passes through a fixed point \((x_1, y_1)\) and has a slope 'm', its equation is given by \(y - y_1 = m(x - x_1)\). This form is incredibly useful when you know a single point on the line and its inclination (slope).

1. Two-Point Form: When a line passes through two distinct points \((x_1, y_1)\) and \((x_2, y_2)\), its equation can be found using the formula \(y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\). This form is essentially a derivation of the point-slope form, where the slope 'm' is calculated using the two given points: \(m = \frac{y_2 - y_1}{x_2 - x_1}\).

1. Slope-Intercept Form: A line with slope 'm' and y-intercept 'c' (the point where the line crosses the y-axis, i.e., \((0, c))\) has the equation \(y = mx + c)\). This form is very intuitive as it directly tells you how steep the line is and where it crosses the y-axis.

1. Intercept Form: If a line makes x-intercept 'a' (meaning it passes through \((a, 0))\) and y-intercept 'b' (meaning it passes through \((0, b))\), its equation is given by \(\frac{x}{a} + \frac{y}{b} = 1\). This form is perfect when you know both points where the line cuts the coordinate axes.

1. Normal Form: This is a slightly more advanced form. The equation of a line whose perpendicular distance from the origin is 'p' and the angle which the normal (perpendicular) to the line makes with the positive x-axis is \(\alpha\) (alpha) is given by \(x \cos \alpha + y \sin \alpha = p\). Remember that 'p' must always be positive. This form is particularly useful in situations involving distances from the origin and angles.

Key Definitions for Straight Lines

Slope (m)

The slope of a straight line is a measure of its steepness and direction. It is defined as the change in y divided by the change in x between two distinct points on the line, i.e., \(m = \frac{\Delta y}{\Delta x}\). A positive slope indicates an upward trend, a negative slope indicates a downward trend, a zero slope means a horizontal line, and an undefined slope means a vertical line.

Intercepts

The points where a line crosses the x-axis and y-axis are called the x-intercept and y-intercept, respectively. The x-intercept has coordinates \((a, 0))\ and the y-intercept has coordinates \((0, b))\, where 'a' and 'b' are the respective intercept values.

Normal to a Line

A normal to a line is another line or segment that is perpendicular to the given line. In the context of the normal form, it refers to the perpendicular drawn from the origin to the line.

Perpendicular Distance (p)

In the normal form of a line, 'p' represents the shortest (perpendicular) distance from the origin \((0, 0))\ to the straight line. This value is always considered positive.

Worked Examples: Applying Line Forms

• Example 1: Using Point-Slope Form Find the equation of a line passing through the point \((2, -3))\ with a slope of \(-\frac{1}{2})\. Step 1: Identify given information. We have a point \((x_1, y_1) = (2, -3))\ and the slope \(m = -\frac{1}{2})\. Step 2: Apply the point-slope formula. The point-slope form is \(y - y_1 = m(x - x_1))\. Substitute the values: \(y - (-3) = -\frac{1}{2}(x - 2))\. Step 3: Simplify the equation. \(y + 3 = -\frac{1}{2}(x - 2))\. Multiply both sides by 2 to remove the fraction: \(2(y + 3) = -1(x - 2))\. \(2y + 6 = -x + 2)\. Rearrange into the standard form \(Ax + By + C = 0)\): \(x + 2y + 6 - 2 = 0)\. \(x + 2y + 4 = 0)\. Final Answer: The equation of the line is \(x + 2y + 4 = 0)\.
• Example 2: Using Two-Point Form and Slope-Intercept Form Find the equation of a line that passes through the points \((1, 2))\ and \((3, 4))\, and express it in slope-intercept form. Step 1: Identify given information. We have two points: \((x_1, y_1) = (1, 2))\ and \((x_2, y_2) = (3, 4))\. Step 2: Calculate the slope (m). \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 2}{3 - 1} = \frac{2}{2} = 1\). Step 3: Use the point-slope form (or two-point form) to find the equation. Using point-slope form with \((1, 2))\ and \(m=1)\): \(y - y_1 = m(x - x_1))\. \(y - 2 = 1(x - 1))\. \(y - 2 = x - 1)\. Step 4: Convert to slope-intercept form (y = mx + c). Rearrange the equation from Step 3: \(y = x - 1 + 2)\. \(y = x + 1\). Final Answer: The equation of the line in slope-intercept form is \(y = x + 1\). Here, the slope \(m=1)\ and the y-intercept \(c=1)\.
• Example 3: Converting a General Equation to Normal Form Convert the equation \(3x + 4y = 10)\ into the normal form. Step 1: Ensure the constant term is on the right side and positive. The equation is \(3x + 4y = 10)\. The constant term is already positive and on the right side. Step 2: Calculate \(\sqrt{A^2 + B^2})\ from the standard form \(Ax + By + C = 0)\ or \(Ax + By = C)\. Here, \(A = 3)\ and \(B = 4)\. So, \(\sqrt{A^2 + B^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\). Step 3: Divide the entire equation by \(\sqrt{A^2 + B^2})\. Divide \(3x + 4y = 10)\ by 5: \(\frac{3x}{5} + \frac{4y}{5} = \frac{10}{5})\. \(\frac{3}{5}x + \frac{4}{5}y = 2\). Step 4: Identify \(\cos \alpha)\ , \(\sin \alpha)\ and 'p'. Comparing this with \(x \cos \alpha + y \sin \alpha = p)\: \(\cos \alpha = \frac{3}{5})\, \(\sin \alpha = \frac{4}{5})\, \(p = 2\). Final Answer: The normal form of the equation is \(\frac{3}{5}x + \frac{4}{5}y = 2\), where the perpendicular distance from the origin is 2 units and the normal makes an angle \(\alpha = \cos^{-1}(\frac{3}{5}))\ (approximately \(53.13^\circ))\ with the positive x-axis.

Exam Tips and Avoiding Common Mistakes

When working with different forms of straight line equations, keep these points in mind to score well:

1. Choosing the Right Form: Always start by identifying what information is given (points, slope, intercepts, or perpendicular distance from origin). This will directly tell you which form is most efficient to use. For example, if two points are given, the two-point form (or calculating slope and then using point-slope) is best.
2. Sign Conventions: Pay close attention to signs, especially when dealing with negative slopes, negative intercepts, or coordinates. A common error is mixing up \((y - y_1))\ with \((y + y_1))\ when \(y_1)\ is negative.
3. Normal Form Nuances: Remember that 'p' (perpendicular distance from the origin) must always be positive. If, after dividing by \(\sqrt{A^2 + B^2})\, the right-hand side is negative, you must multiply the entire equation by -1 to make 'p' positive, which will also change the signs of the \(x \cos \alpha)\ and \(y \sin \alpha)\ terms. Also, \(\alpha)\ is the angle the normal makes with the x-axis, not the line itself.
4. Verification: After finding the equation, a quick check can save marks. If you used a point in your calculation, plug it back into your final equation to see if it satisfies it. If you found intercepts, check if the equation yields those intercepts when \(x=0)\ or \(y=0)\.

Practice Questions with Solutions

• Q: Find the equation of the line passing through \((1, -2))\ and having a slope of 3. A: Step 1: Identify the given point \((x_1, y_1) = (1, -2))\ and slope \(m = 3)\. Step 2: Use the point-slope form: \(y - y_1 = m(x - x_1))\. Step 3: Substitute the values: \(y - (-2) = 3(x - 1))\. Step 4: Simplify the equation: \(y + 2 = 3x - 3\). Step 5: Rearrange to standard form: \(3x - y - 3 - 2 = 0 \implies 3x - y - 5 = 0\). Final answer: The equation of the line is \(3x - y - 5 = 0\).
• Q: Determine the equation of the line that cuts off intercepts 4 and -3 on the x and y axes respectively. A: Step 1: Identify the x-intercept \(a = 4)\ and y-intercept \(b = -3)\. Step 2: Use the intercept form: \(\frac{x}{a} + \frac{y}{b} = 1\). Step 3: Substitute the values: \(\frac{x}{4} + \frac{y}{-3} = 1\). Step 4: Find a common denominator (LCM of 4 and 3 is 12) and simplify: \(\frac{3x - 4y}{12} = 1\). Step 5: Multiply both sides by 12: \(3x - 4y = 12\). Step 6: Rearrange to standard form: \(3x - 4y - 12 = 0\). Final answer: The equation of the line is \(3x - 4y - 12 = 0\).
• Q: A line passes through the points \((0, -2))\ and \((5, 3))\. Find its equation and express it in slope-intercept form. A: Step 1: Identify the two given points: \((x_1, y_1) = (0, -2))\ and \((x_2, y_2) = (5, 3))\. Step 2: Calculate the slope \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-2)}{5 - 0} = \frac{3 + 2}{5} = \frac{5}{5} = 1\). Step 3: Use the point-slope form with \((0, -2))\ and \(m=1)\: \(y - (-2) = 1(x - 0))\. Step 4: Simplify: \(y + 2 = x\). Step 5: Express in slope-intercept form \(y = mx + c)\: \(y = x - 2\). Final answer: The equation of the line in slope-intercept form is \(y = x - 2\).
• Q: Find the normal form of the line \(x - \sqrt{3}y + 8 = 0\). A: Step 1: Rewrite the equation to have the constant term positive on the right side: \(x - \sqrt{3}y = -8\). Step 2: Since the right-hand side is negative, multiply the entire equation by -1 to make 'p' positive: \(-x + \sqrt{3}y = 8\). Step 3: Calculate \(\sqrt{A^2 + B^2})\ for \(-x + \sqrt{3}y = 8\) where \(A = -1)\ and \(B = \sqrt{3})\. \(\sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2\). Step 4: Divide the modified equation by 2: \(\frac{-x}{2} + \frac{\sqrt{3}y}{2} = \frac{8}{2})\. Step 5: Simplify: \(-\frac{1}{2}x + \frac{\sqrt{3}}{2}y = 4\). Final answer: The normal form of the line is \(-\frac{1}{2}x + \frac{\sqrt{3}}{2}y = 4\). Here, \(\cos \alpha = -\frac{1}{2})\, \(\sin \alpha = \frac{\sqrt{3}}{2})\, which means \(\alpha = 120^\circ)\ and \(p = 4\).

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