The trajectory equation of the point with equal distance to the line y = 0 and the line y = √ 3 (x + 1), The answer is (√ 3x-3y + √ 3) (√ 3x + y + √ 3) = 0 I'm dull and love to get to the top of things, so the more detailed the better,

The trajectory equation of the point with equal distance to the line y = 0 and the line y = √ 3 (x + 1), The answer is (√ 3x-3y + √ 3) (√ 3x + y + √ 3) = 0 I'm dull and love to get to the top of things, so the more detailed the better,

Let the coordinates of points with equal distances be p (x, y)
So the distance from P to the line y = 0 is | y|
The distance from P to the line y = radical 3 (x + 1) is d = | radical 3x-y + radical 3 | / radical (3 + 1)
So we have | y | = | root 3x-y + root 3 | / 2
That is: 2Y = root 3x-y + root 3 or - 2Y = root 3x-y + root 3
That is, there is root 3x-3y + root 3 = 0 or root 3x + y + root 3 = 0
That is, (√ 3x-3y + √ 3) (√ 3x + y + √ 3) = 0