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 .
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