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About LinkBacksFind the vector perpendicular to the line 3x-5y+2=0.
Write in the form y = mx + c, i.e. y = (3/5)x + 2/5, which has gradient = 3/5.Originally Posted by johnsy123
The gradients of perpendicular lines have a product of -1. So the gradient of your perpendicular is -5/3.
So any line in the family of lines y = (-5/3)x + c will be perpendicular. For simplicity we'll just consider y = (-5/3)x.
If you note from the slope that this perpendicular line moves down 5 units as you move to the right 3 units, then you can write this line in its vector form t(3i - 5j), where t is an arbitrary constant. If you know a point that this line goes through, say (x_0, y_0), then the line can be written as (x_0 + 3t)i + (y_0 - 5t)j.
If the vector (x, y) denotes the staionary vector of any point on the given line then the equation of the line is the result of a scalar product:
(3, -5) . (x, y) = -2
where the vector (3, -5) is the normal vector, that means it is perpendicular to the line.
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