MP Board · Class 11 · Mathematics · Straight LinesReduce the equation $\sqrt{3}x + y - 8 = 0$ into normal form $x \cos \omega + y \sin \omega = p$. What is the value of $p$?
To reduce $Ax + By + C = 0$ to normal form, divide by $\sqrt{A^2 + B^2}$. Here $A = \sqrt{3}, B = 1, C = -8$. $\sqrt{A^2 + B^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$. Dividing the equation by 2: $\frac{\sqrt{3}}{2}x + \frac{1}{2}y - 4 = 0 \implies \frac{\sqrt{3}}{2}x + \frac{1}{2}y = 4$. Comparing with $x \cos \omega + y \sin \omega = p$, we get $p = 4$. Therefore, the correct option is B.
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Straight Lines.
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Straight Lines.
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Straight Lines.