find the equation of the line which satisfy the given condition: Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30o.

Class 11 | Exercise 10.2 | Maths | NCERT | Straight Lines

Given: \[\mathbf{p}\text{ }=\text{ }\mathbf{5}\] and \[\mathbf{\omega }\text{ }=\text{ }\mathbf{30}{}^\circ \]

We realize that the condition of the line having typical distance p from the beginning and point ω which the ordinary makes with the positive heading of x-axis is given by \[\mathbf{x}\text{ }\mathbf{cos}\text{ }\mathbf{\omega }\text{ }+\text{ }\mathbf{y}\text{ }\mathbf{sin}\text{ }\mathbf{\omega }\text{ }=\text{ }\mathbf{p}.\]

Subbing the qualities in the situation, we get

\[\mathbf{x}\text{ }\mathbf{cos30}{}^\circ \text{ }+\text{ }\mathbf{y}\text{ }\mathbf{sin30}{}^\circ \text{ }=\text{ }\mathbf{5}\]

\[\mathbf{x}\left( \surd \mathbf{3}/\mathbf{2} \right)\text{ }+\text{ }\mathbf{y}\left( \text{ }\mathbf{1}/\mathbf{2}\text{ } \right)\text{ }=\text{ }\mathbf{5}\]

\[\surd \mathbf{3}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{5}\left( \mathbf{2} \right)\text{ }=\text{ }\mathbf{10}~~~~~~~~\]

\[\surd \mathbf{3}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{y}\text{ }\text{ }\mathbf{-10}\text{ }=\text{ }\mathbf{0}\]

∴ The condition of the line is \[\surd \mathbf{3}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{y}\text{ }\text{ }\mathbf{-10}\text{ }=\text{ }\mathbf{0}.\]

i

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