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A line is perpendicular to 3xy+11=0 and passes th...

A line is perpendicular to 3xy+11=0 and passes through the point (1, -5). Find its equation.

A.

3y - x -14 = 0

B.

3x + y + 1 = 0

C.

3y + x + 1 = 0

D.

3y + x + 14 = 0

Correct answer is D

3xy+11=0y=3x+11

Gradient=3

Gradient of perpendicular line = 13

3(y + 5) = 1 - x

3y + x + 15 - 1 = 3y + x + 14 = 0