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Straight Lines - General Equation of a Line

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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General Equation of a Line: Every first-degree equation in two variables xx and yy of the form Ax+By+C=0Ax + By + C = 0 (where AA and BB are not both zero) represents a straight line. Visually, this is a continuous straight path on a 2D Cartesian plane that extends infinitely.

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Reduction to Slope-Intercept Form: The general equation Ax+By+C=0Ax + By + C = 0 can be transformed into y=mx+cy = mx + c by isolating yy, giving y=(−AB)x+(−CB)y = (-\frac{A}{B})x + (-\frac{C}{B}). In this form, the coefficient of xx represents the slope (steepness) and the constant term represents the y-intercept (where the line crosses the vertical axis).

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Reduction to Intercept Form: If C≠0C \neq 0, the general equation can be rewritten as xa+yb=1\frac{x}{a} + \frac{y}{b} = 1, where a=−CAa = -\frac{C}{A} and b=−CBb = -\frac{C}{B}. Geometrically, aa and bb are the distances from the origin to the points where the line intersects the x-axis and y-axis respectively.

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Reduction to Normal Form: The general equation can be expressed as xcos⁡α+ysin⁡α=px \cos \alpha + y \sin \alpha = p. Visually, pp is the length of the perpendicular segment (the normal) drawn from the origin (0,0)(0,0) to the line, and α\alpha is the angle this normal makes with the positive direction of the x-axis.

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Distance of a Point from a Line: The perpendicular distance dd from a point (x1,y1)(x_1, y_1) to the line Ax+By+C=0Ax + By + C = 0 is the shortest distance between them. Visually, this is represented by a line segment starting at the point and meeting the line at a 90∘90^{\circ} angle.

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Distance Between Parallel Lines: Two lines are parallel if they have the same slope, meaning their equations can be written as Ax+By+C1=0Ax + By + C_1 = 0 and Ax+By+C2=0Ax + By + C_2 = 0. The distance between them is the constant length of any perpendicular segment connecting the two lines, showing how far apart they are spaced.

📐Formulae

General Form: Ax+By+C=0Ax + By + C = 0

Slope (mm): m=−ABm = -\frac{A}{B}

Y-intercept (cc): c=−CBc = -\frac{C}{B}

X-intercept (aa): a=−CAa = -\frac{C}{A}

Normal Form conversion: cos⁡α=±AA2+B2\cos \alpha = \pm \frac{A}{\sqrt{A^2+B^2}}, sin⁡α=±BA2+B2\sin \alpha = \pm \frac{B}{\sqrt{A^2+B^2}}, p=∣C∣A2+B2p = \frac{|C|}{\sqrt{A^2+B^2}}

Distance from (x1,y1)(x_1, y_1) to Ax+By+C=0Ax + By + C = 0: d=∣Ax1+By1+C∣A2+B2d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

Distance between parallel lines Ax+By+C1=0Ax + By + C_1 = 0 and Ax+By+C2=0Ax + By + C_2 = 0: d=∣C1−C2∣A2+B2d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

💡Examples

Problem 1:

Reduce the equation 3x−4y+10=03x - 4y + 10 = 0 to slope-intercept form and find its slope and y-intercept.

Solution:

  1. Given the equation: 3x−4y+10=03x - 4y + 10 = 0. \ 2. Isolate the yy term: −4y=−3x−10-4y = -3x - 10. \ 3. Divide by −4-4: y=−3−4x−10−4y = \frac{-3}{-4}x - \frac{10}{-4}. \ 4. Simplify: y=34x+52y = \frac{3}{4}x + \frac{5}{2}. \ 5. Comparing with y=mx+cy = mx + c, we get m=34m = \frac{3}{4} and c=52c = \frac{5}{2}.

Explanation:

To find the slope and y-intercept, we rearrange the general linear equation into the slope-intercept form y=mx+cy = mx + c by solving for yy.

Problem 2:

Find the distance between the parallel lines 3x−4y+7=03x - 4y + 7 = 0 and 3x−4y+5=03x - 4y + 5 = 0.

Solution:

  1. Identify coefficients: A=3,B=−4,C1=7,C2=5A = 3, B = -4, C_1 = 7, C_2 = 5. \ 2. Use the formula for distance between parallel lines: d=∣C1−C2∣A2+B2d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}. \ 3. Substitute the values: d=∣7−5∣32+(−4)2d = \frac{|7 - 5|}{\sqrt{3^2 + (-4)^2}}. \ 4. Calculate the denominator: 9+16=25=5\sqrt{9 + 16} = \sqrt{25} = 5. \ 5. Calculate the numerator: ∣2∣=2|2| = 2. \ 6. Result: d=25d = \frac{2}{5} units.

Explanation:

Since the coefficients of xx and yy are identical in both equations, the lines are parallel. We apply the specific formula for the distance between parallel lines using their constant terms.