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Straight Lines - Normal Form of a Line

Grade 11CBSE

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

🔑Concepts

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The Normal Form of a straight line is a way to define a line's position using the length of the perpendicular (normal) drawn from the origin (0,0)(0,0) to the line and the angle this perpendicular makes with the positive x-axis. Visually, imagine a line passing through any quadrant; the 'normal' is the unique shortest line segment connecting the origin to that line at a 90∘90^{\circ} angle.

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The parameter pp represents the length of the normal from the origin to the line. By definition, distance is non-negative, so pp must always be greater than or equal to zero (p≥0p \ge 0). On a graph, pp is the radius of a circle centered at the origin that is tangent to the line.

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The parameter ω\omega (omega) is the angle that the normal makes with the positive direction of the x-axis, measured in the counter-clockwise direction. The range of this angle is 0≤ω<360∘0 \le \omega < 360^{\circ} (or 2π2\pi radians). The value of ω\omega determines which direction the normal points, which in turn determines the slope of the line itself.

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The relationship between the line and its normal is perpendicular. Because the normal has an angle ω\omega, its slope is tan⁡ω\tan \omega. Since the line is perpendicular to the normal, the slope of the line is m=−1tan⁡ω=−cot⁡ωm = -\frac{1}{\tan \omega} = -\cot \omega.

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To convert a general equation Ax+By+C=0Ax + By + C = 0 into normal form, you must first move the constant term to the right side: Ax+By=−CAx + By = -C. If −C-C is negative, multiply the entire equation by −1-1 to ensure the right side is positive, as it represents the distance pp.

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After ensuring the constant term is positive, divide every term in the equation by A2+B2\sqrt{A^2 + B^2}. This results in the coefficients of xx and yy becoming cos⁡ω\cos \omega and sin⁡ω\sin \omega respectively, while the constant on the right becomes pp.

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Identifying the quadrant of ω\omega is crucial. If the coefficient of xx (cos⁡ω\cos \omega) is negative and the coefficient of yy (sin⁡ω\sin \omega) is positive, the normal lies in the second quadrant. If both are negative, it lies in the third quadrant, and so on.

📐Formulae

Standard Normal Form: xcos⁡ω+ysin⁡ω=px \cos \omega + y \sin \omega = p

Normal length from origin: p=∣C∣A2+B2p = \frac{|C|}{\sqrt{A^2 + B^2}}

Relationship for angle: cos⁡ω=−A±A2+B2\cos \omega = \frac{-A}{\pm\sqrt{A^2+B^2}} and sin⁡ω=−B±A2+B2\sin \omega = \frac{-B}{\pm\sqrt{A^2+B^2}}

Condition for pp: p>0p > 0

Slope of the line in terms of ω\omega: m=−cot⁡ωm = -\cot \omega

💡Examples

Problem 1:

Find the equation of the line for which the length of the perpendicular from the origin is 55 units and the angle which the perpendicular makes with the positive x-axis is 30∘30^{\circ}.

Solution:

Step 1: Identify the given values. We have p=5p = 5 and ω=30∘\omega = 30^{\circ}. Step 2: Use the Normal Form equation xcos⁡ω+ysin⁡ω=px \cos \omega + y \sin \omega = p. Step 3: Substitute the values: xcos⁡30∘+ysin⁡30∘=5x \cos 30^{\circ} + y \sin 30^{\circ} = 5. Step 4: Evaluate trigonometric ratios: cos⁡30∘=32\cos 30^{\circ} = \frac{\sqrt{3}}{2} and sin⁡30∘=12\sin 30^{\circ} = \frac{1}{2}. Step 5: Substitute these back into the equation: x(32)+y(12)=5x \left(\frac{\sqrt{3}}{2}\right) + y \left(\frac{1}{2}\right) = 5. Step 6: Simplify by multiplying the entire equation by 22: 3x+y=10\sqrt{3}x + y = 10.

Explanation:

This is a direct application of the normal form. We simply plug the distance pp and the angle ω\omega into the standard equation and simplify.

Problem 2:

Reduce the equation 3x+y−8=0\sqrt{3}x + y - 8 = 0 into normal form. Find the values of pp and ω\omega.

Solution:

Step 1: Rewrite the equation as 3x+y=8\sqrt{3}x + y = 8. Here, the constant term on the right is already positive (8>08 > 0). Step 2: Calculate A2+B2\sqrt{A^2 + B^2}, where A=3A = \sqrt{3} and B=1B = 1. So, (3)2+12=3+1=4=2\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2. Step 3: Divide the entire equation by 22: 32x+12y=82\frac{\sqrt{3}}{2}x + \frac{1}{2}y = \frac{8}{2}, which simplifies to 32x+12y=4\frac{\sqrt{3}}{2}x + \frac{1}{2}y = 4. Step 4: Compare with xcos⁡ω+ysin⁡ω=px \cos \omega + y \sin \omega = p. We find p=4p = 4. Step 5: Determine ω\omega from cos⁡ω=32\cos \omega = \frac{\sqrt{3}}{2} and sin⁡ω=12\sin \omega = \frac{1}{2}. Both are positive, so ω\omega is in the first quadrant. ω=30∘\omega = 30^{\circ} or π6\frac{\pi}{6}.

Explanation:

To reduce a general equation to normal form, we divide by the magnitude of the coefficients' vector A2+B2\sqrt{A^2+B^2}. This normalizes the coefficients so they represent the sine and cosine of the same angle.