since the line √3x+y-1=0 makes angle of -30° with the x-axis, the desired line is inclined at an angle of +30°. So, we have
y+2 = 1/√3 (x-3)
y = 1/√3 x - (2+√3)
or, as the other line is given is standard form,
x-√3y-(3+2√3) = 0
Find the equation of the straight line passing through the point(3,-2) and making an angle of 60° with the line√3x+y-1=0
Plzz help someone
1 answer
1 answer