Find the equation of the straight line passing through (3,-2) and making an angle of 60 degree with the positive direction of y-axis
To find the equation of a straight line passing through a given point and making an angle with the positive direction of the y-axis, we need to first determine the slope of the line.
The slope of a line can be found using trigonometry. Since the line makes an angle of 60 degrees with the positive direction of the y-axis, the complementary angle between the line and the x-axis is 30 degrees (90 degrees - 60 degrees).
The slope of the line can be calculated using the tangent of the angle. The tangent of 30 degrees is √3/3.
Next, we can use the point-slope form of a linear equation, which is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) are the coordinates of a point on the line.
Given that the point (3, -2) lies on the line, we can substitute it into the point-slope form equation:
y - (-2) = (√3/3)(x - 3)
Simplifying this equation, we have:
y + 2 = (√3/3)(x - 3)
Multiplying (√3/3) by (x - 3), we get:
y + 2 = (√3/3)x - (√3/3)(3)
y + 2 = (√3/3)x - √3
Rearranging this equation in the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept, we get:
y = (√3/3)x - √3 - 2
Finally, we can simplify the equation to its standard form:
√3x - 3y = 3√3 + 6
Therefore, the equation of the straight line passing through (3, -2) and making an angle of 60 degrees with the positive direction of the y-axis is √3x - 3y = 3√3 + 6.
The slope of the line is the tan of the angle that the line makes with the x-axis
so if the line makes and angle of 60° with the y-axis , it will make an angle of 30° with the x-axis
so we have tan30° = 1/√3
equation:
y + 2 = (1/√3)(x-3)
√3y + 2√3 = x - 3
x - √3y = 2√3 + 3
change to whatever form of the equation that is needed.