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Home » Find the equation of a line making an angle of 150° with the x–axis and cutting off an intercept 2 from y–axis.
Find the equation of a line making an angle of 150° with the x–axis and cutting off an intercept 2 from y–axis.
CBSE | Class 11 | Exercise 22.3 | Maths | RD Sharma | Straight Lines
Find the equation of a line making an angle of 150° with the x–axis and cutting off an intercept 2 from y–axis.

Given: A line which makes an angle of

    \[{{150}^{o}}~\]

with the x–axis and cutting off an intercept at

    \[2\]

By using the formula,

The equation of a line is

    \[y\text{ }=\text{ }mx\text{ }+\text{ }c\]

We know that angle, 

    \[\theta \text{ }=\text{ }{{150}^{o}}\]

The slope of the line,

    \[m\text{ }=\text{ }tan\text{ }\theta \]

Where,

    \[m\text{ }=\text{ }tan\text{ }{{150}^{o}}\]

    \[=\text{ }-1/~\surd 3\]

Coordinate of y–intercept is

    \[\left( 0,\text{ }2 \right)\]

The required equation of the line is

    \[y\text{ }=\text{ }mx\text{ }+\text{ }c\]

Now substitute the values, we get

    \[y\text{ }=\text{ }-x/\surd 3\text{ }+\text{ }2\]

    \[\surd 3y-2\surd 3\text{ }+\text{ }x\text{ }=\text{ }0\]

So,

    \[x\text{ }+~\surd 3y\text{ }=\text{ }2\surd 3\]

∴ The equation of line is

    \[x\text{ }+~\surd 3y\text{ }=\text{ }2\surd 3\]

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