(3,10) perendicular to x-3y=-4

Find the equation of the line that passes through the point (3,10) and ios perpenicular to the line x-3y=-4
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1 Answer

x-3y=-4 can be written y=⅓x+4/3, which has gradient ⅓. The gradient of the perpendicular is therefore -3.

The equation of the perpendicular passing through (3,10) is:

y-10=-3(x-3), y=10-3x+9=-3x+19, also written 3x+y=19.

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