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Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with the positive direction of x–axis is 15°.
CBSE | Class 11 | Exercise 23.7 | Maths | RD Sharma | Straight Lines
Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with the positive direction of x–axis is 15°.

Given:

    \[p\text{ }=\text{ }4,\text{ }\alpha \text{ }=\text{ }15{}^\circ \]

The equation of the line in normal form is given by

as,

    \[~cos\text{ }15{}^\circ ~=\text{ }cos\text{ }\left( 45{}^\circ ~\text{ }30{}^\circ \right)\]

    \[=\text{ }cos45{}^\circ cos30{}^\circ ~+\text{ }sin45{}^\circ sin30{}^\circ \]

And

    \[Cos\text{ }\left( A\text{ }\text{ }B \right)\]

    \[~=\text{ }cos\text{ }A\text{ }cos\text{ }B\text{ }+\text{ }sin\text{ }A\text{ }sin\text{ }B\]

Hence ,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 25

And

    \[sin\text{ }15\text{ }=\text{ }sin\text{ }\left( 45{}^\circ ~\text{ }30{}^\circ \right)\]

    \[~=\text{ }sin\text{ }45{}^\circ ~cos\text{ }30{}^\circ ~\text{ }cos\text{ }45{}^\circ ~sin\text{ }30{}^\circ \]

    \[Sin\text{ }\left( A\text{ }\text{ }B \right)\]

    \[=\text{ }sin\text{ }A\text{ }cos\text{ }B\text{ }\text{ }cos\text{ }A\text{ }sin\text{ }B\]

So,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 26

Now, on using the formula,

    \[x\text{ }cos~\alpha \text{ }+\text{ }y\text{ }sin\text{ }\alpha \text{ }=\text{ }p\]

Now, putting the values, we have

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 27

    \[\left( \surd 3+1 \right)x\text{ }+\left( \surd 3-1 \right)\text{ }y\text{ }=\text{ }8\surd 2\]

∴ The equation of line in normal form is:

    \[~\left( \surd 3+1 \right)x\text{ }+\left( \surd 3-1 \right)\text{ }y\text{ }=\text{ }8\surd 2.\]

i

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