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zhannawk [14.2K]
3 years ago
12

Here's a question ~

Mathematics
2 answers:
spayn [35]3 years ago
8 0

<u>kindly refer to the attachment below !!!</u>

finlep [7]3 years ago
3 0

Answer:

\sf\longmapsto \:  ( \sqrt{3}   - 2)x + (2 \sqrt{3 }  + 1)y =  - 1 + 8 \sqrt{3}

Step-by-step explanation:

It is given that –

Slope of the first line is–

\sf\longmapsto \:  m_{1} = 2

Let,

the slope of the Another line be –

\sf\longmapsto \: m _{2}

Now,

The angle between the two lines is 60°.

Let's start solving!

\sf\longmapsto \:  \tan(60°)  | \frac{ m_{1} - m_ {2} }{1 + m_{1} m_ {2}} |

\sf\longmapsto \:  \sqrt{3}  =   | \frac{2 -m_{2} }{1 + 2m_{2}} |

\sf\longmapsto \:  \sqrt{3}  = ± \: \: ( \frac{2 -m_{2} }{1 + 2m_{2}} )

\sf\longmapsto \:  \sqrt{3}  =  \frac{2 -m_{2} }{1 + 2m_{2}}

\sf\longmapsto \:  \sqrt{3} (1 + 2m_{2}) = 2 - m_{2}

\sf\longmapsto \:  \sqrt{3}  + 2  \sqrt{3} m_{2} + m_{2} = 2

\sf\longmapsto \: m_{2} =  \frac{2 -  \sqrt{3} }{(2 \sqrt{3} + 1) }

The equation of line passing through point (2,3) and having a slope of –

\sf\longmapsto \: m_{2} =  \frac{(2 -  \sqrt{3} )}{2 \sqrt{3} + 1 }

is–

\sf\longmapsto \: (y - 3) =  \frac{2 -  \sqrt{3} }{2 \sqrt{3} + 1 } (x - 2)

\sf\longmapsto \: ( 2\sqrt{3}  + 1)y - 3(2 \sqrt{3}  + 1) = (2  - \sqrt{3} )x - 2(2 -  \sqrt{3}

\sf\longmapsto \:  ( \sqrt{3}   - 2)x + (2 \sqrt{3 }  + 1)y =  - 1 + 8 \sqrt{3}

Hence the equation of the other line is -

\sf\longmapsto \:  ( \sqrt{3}   - 2)x + (2 \sqrt{3 }  + 1)y =  - 1 + 8 \sqrt{3}

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