# Find the equation of the line which passes through the point (-2,3) and makes an angle 60 with the positive direction of x axis .

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The equation for the lines is:

y-y1 = m(x-x1) where m is the slope

Let us substitute with (-2,3)

==> y- 3 = m(x+2)

Since the line makes a 60 degree angle , then

the slope = y2-y1/x2-x1 = tan60 = sqrt3

==> y- 3 = (sqrt3)(x+2)

==> y-3 = sqrt3)*x + 2sqrt3

==> y= (sqrt3)*x + (3+sqrt3)

For solving the problem, we'll use the next working rule: the equation of the line which passes through a given point and has an inclination "a" is:

(**x-x1) / cos a = (y-y1) / sin a = r**,

where (x1,y1) are the coordinates of the given point and r is the distance between (x,y) and (x1,y1).

In our case, the coordinates of the given point (-2,3) and the inclination a = 60 degrees.

The equation is:

[x-(-2)]/cos 60 = (y-3)/ sin 60 = r

cos 60 = 1/2 and sin 60 = sqrt3/2

(x+2)/(1/2) = (y-3)/(sqrt3/2)

We'll divide by 2:

**x+2 = (y-3)/sqrt 3 = r/2 **

Let the equation of the line be y = mx+c, the slope intercept form.

Since the line passes through (2,3), the equation is (y-3) = mx(x-2).

Since the line is inclined 60 degree , the slope of the line m = tan 60 = sqrt3.

Therefore the equation of the line is (y-3) = (tan60)(x-2)

y-3 = sqrt3(x-2)

y =( sqrt3)x -2sqrt3 +3