Answer:
[tex]\sf\longmapsto \: ( \sqrt{3} - 2)x + (2 \sqrt{3 } + 1)y = - 1 + 8 \sqrt{3} [/tex]
Step-by-step explanation:
It is given that –
Slope of the first line is–
[tex]\sf\longmapsto \: m_{1} = 2[/tex]
Let,
the slope of the Another line be –
[tex]\sf\longmapsto \: m _{2}[/tex]
Now,
The angle between the two lines is 60°.
Let's start solving!
[tex]\sf\longmapsto \: \tan(60°) | \frac{ m_{1} - m_ {2} }{1 + m_{1} m_ {2}} | [/tex]
[tex]\sf\longmapsto \: \sqrt{3} = | \frac{2 -m_{2} }{1 + 2m_{2}} | [/tex]
[tex]\sf\longmapsto \: \sqrt{3} = ± \: \: ( \frac{2 -m_{2} }{1 + 2m_{2}} )[/tex]
[tex]\sf\longmapsto \: \sqrt{3} = \frac{2 -m_{2} }{1 + 2m_{2}} [/tex]
[tex]\sf\longmapsto \: \sqrt{3} (1 + 2m_{2}) = 2 - m_{2}[/tex]
[tex]\sf\longmapsto \: \sqrt{3} + 2 \sqrt{3} m_{2} + m_{2} = 2[/tex]
[tex]\sf\longmapsto \: m_{2} = \frac{2 - \sqrt{3} }{(2 \sqrt{3} + 1) } [/tex]
The equation of line passing through point (2,3) and having a slope of –
[tex]\sf\longmapsto \: m_{2} = \frac{(2 - \sqrt{3} )}{2 \sqrt{3} + 1 } [/tex]
is–
[tex]\sf\longmapsto \: (y - 3) = \frac{2 - \sqrt{3} }{2 \sqrt{3} + 1 } (x - 2)[/tex]
[tex]\sf\longmapsto \: ( 2\sqrt{3} + 1)y - 3(2 \sqrt{3} + 1) = (2 - \sqrt{3} )x - 2(2 - \sqrt{3} [/tex]
[tex]\sf\longmapsto \: ( \sqrt{3} - 2)x + (2 \sqrt{3 } + 1)y = - 1 + 8 \sqrt{3} [/tex]
Hence the equation of the other line is -
[tex]\sf\longmapsto \: ( \sqrt{3} - 2)x + (2 \sqrt{3 } + 1)y = - 1 + 8 \sqrt{3} [/tex]
