Hi, Im really confused with this question. Please explain very clearly thanks! :)
Find the equation of the straight line which passes through the point (2, 3) and is inclined at 30 degrees to the positive direction of the x-axis.
Find the equation of the straight line which passes through the point (2, 3) and is inclined at 30 degrees to the positive direction of the x-axis.
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"inclined at 30 degrees" means the slope (or tangent) of the line is m = tan(30) = sqrt(3)/3. The slope is the tangent value.
The point slope formula is y = mx + b.
Given the point (2,3) and knowing the slope is sqrt(3)/3, substituting into the equation to find b,
3 = (sqrt(3)/3) (2) + b
b = 3 - 2sqrt(3)/3
So the equation of the line is
y = (sqrt(3)/3)x + 3 - 2sqrt(3)/3
The point slope formula is y = mx + b.
Given the point (2,3) and knowing the slope is sqrt(3)/3, substituting into the equation to find b,
3 = (sqrt(3)/3) (2) + b
b = 3 - 2sqrt(3)/3
So the equation of the line is
y = (sqrt(3)/3)x + 3 - 2sqrt(3)/3