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If p be the length of the perpendicular from the origin on the
straight line $\mathrm{ax}+\mathrm{by}=\mathrm{p}$ and $\mathrm{b}=\frac{\sqrt{3}}{2}$, then what is the angle between the perpendicular and the positive direction of $\begin{array}{ll}\text { x-axis? } & {}\end{array}$
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straight line $\mathrm{ax}+\mathrm{by}=\mathrm{p}$ and $\mathrm{b}=\frac{\sqrt{3}}{2}$, then what is the angle between the perpendicular and the positive direction of $\begin{array}{ll}\text { x-axis? } & {}\end{array}$
Solution:
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Verified Answer
The correct answer is:
$60^{\circ}$
Equation of line is $a x+b y-p=0$, then length of perpendicular, from the origin.
$p=\left|\frac{a \times 0+b \times 0-p}{\sqrt{a^{2}+b^{2}}}\right|$ or $p=\left|\frac{-p}{\sqrt{a^{2}+b^{2}}}\right|$
or $\left|\frac{1}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\right|=1$ or $\mathrm{a}^{2}+\mathrm{b}^{2}=1$
$\mathrm{b}=\frac{\sqrt{3}}{2}$ or $\mathrm{b}^{2}=\frac{3}{4}$
$a^{2}+\frac{3}{4}=1$
$\mathrm{a}^{2}=\frac{1}{4} \Rightarrow \mathrm{a}=\frac{1}{2} \quad\left[\mathrm{a}=-\frac{1}{2}\right.$ not taken since
angle is with $+$ ve direction to $\mathrm{x}$ -axis. $]$ Equation is $\frac{1}{2} x+\frac{\sqrt{3}}{2} y=p$ or $x \cos 60^{\circ}+y \sin 60^{\circ}=p$
Angle $=60^{\circ}$
$p=\left|\frac{a \times 0+b \times 0-p}{\sqrt{a^{2}+b^{2}}}\right|$ or $p=\left|\frac{-p}{\sqrt{a^{2}+b^{2}}}\right|$
or $\left|\frac{1}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\right|=1$ or $\mathrm{a}^{2}+\mathrm{b}^{2}=1$
$\mathrm{b}=\frac{\sqrt{3}}{2}$ or $\mathrm{b}^{2}=\frac{3}{4}$
$a^{2}+\frac{3}{4}=1$
$\mathrm{a}^{2}=\frac{1}{4} \Rightarrow \mathrm{a}=\frac{1}{2} \quad\left[\mathrm{a}=-\frac{1}{2}\right.$ not taken since
angle is with $+$ ve direction to $\mathrm{x}$ -axis. $]$ Equation is $\frac{1}{2} x+\frac{\sqrt{3}}{2} y=p$ or $x \cos 60^{\circ}+y \sin 60^{\circ}=p$
Angle $=60^{\circ}$
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