Answer to Question #91553 in Analytic Geometry for Ra

Question

Find the equation of the line which passes through the point (1,√3) and makes an angle of 30º with the line x-√(3)y+√3=0

Answer 1

Let Slope of the given line is m₁

$x-\sqrt{\smash[b]{3}}y+\sqrt{\smash[b]{3}}=0$

This gives,

$y=x/\sqrt{\smash[b]{3}}+1$,

y=x/√3+1,

where m₁=1/√3=0.577.

Let the slope of the required line which makes an angle 30 degree with the above line is m.

Thus,

$tan30^0=\frac{0.577-m}{1+0.577}$

The value of the left side with tan30⁰=+,-0.577

​Take +0.577.

0.577=(0.577-m)/(1+0.577m);

0.577=(0.577−m)/(1+0.577m);

0.577(1+0.577m)=(0.577-m);

0.577+0.33m=0.577-m.

Thus, m=0.

The line is passing through the point (1,√3).

Thus, the equation of line is y-√3=0.

​Take -0.577.

-0.577=(0.577-m)/(1+0.577m);

-0.577=(0.577−m)/(1+0.577m);

-0.577(1+0.577m)=(0.577-m);

-0.577-0.33m=0.577-m;

-0.577-0.577=-m+0.33m;

-1.154=-0.67m;

m=1.72.

Line is passing through the point (1,√3).

Thus, the equation of the required line is

(y-√3)=1.72(x-1).