Find me equation of the line that is perpendicular to y=-6x +3 and passes through the point (8,-3)

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The gradient of the perpendicular is ⅙ being -(1/(-6)), so its equation is of the form y=x/6+a where a is a constant. Plug in the given point to find a: -3=8/6+a, a=-13/3 so y=x/6-13/3 or 6y=x-26 or x-6y-26=0. All these equations are equivalent.

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