How do you find an equation of the line perpendicular to the graph of 15x-5y=7 that passes through the point at (0,-4)?

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Answer 1 · 2018-03-02T12:51:20

The equation of the perpendicular line is #x+3y=-12#

#15x-5y=7 or 5y= 15x-7 or y= 3x -7/5#

Slope of the line # y= 3x -7/5 [y =mx+c]# is #m_1=3#

The product of slopes of the perpendicular lines is #m_1*m_2=-1#

#:.m_2=-1/3=-1/3#. The equation of the perpendicular line

passing through #(0,-4)# having slope of #m_2=-1/3# is

#y-y_1=m_2(x-x_1) or y-(-4)=-1/3(x-0)# or

#y+4 = -1/3x or y = -1/3x-4 or x+3y=-12# [Ans]