What is the equation of the line with slope # m= -31/36 # that passes through # (-5/6, 13/18) #?

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Answer 1 · 2018-03-09T06:36:23

#216y+186x=1#

Slope of a line #(m) = (y_1-y_2)/(x_1-x_2)# ----(1)

Here , #m=-31/36#

#x_1=x#

#x_2=-5/6#

#y_1=y#

#y_2=13/18#

Put these values in equation(1)

#=> -31/36=(y-13/18)/(x-(-5/6))#

#=> -31/36 = ((18y-13)/cancel18^3)/((6x+5)/cancel6#

#=> -31/cancel36^12=(18y-13)/(cancel3(6x+5)#

Cross-multiply

#=> -31(6x+5)=12(18y-13)#

#=> -186x-155=216y-156#

#=> 156-155=216y+186x#

#=> 1=216y+186x#

Answer 2 · 2018-03-09T06:53:24

#color(orange)(186x + 216y = 1#

Given slope and a point on the line, we can write the equation using

#(y - y_1 ) = m (x - x_1)#

where m is the slope and #(x_1, y_1)# the coordinates of the point.

Hence the equation is

#y - (13/18) = -(31/36) * (x + 5/6)#

#y = -(31/36)x - (31/36)*(5/6) + 13 / 18#

#y = [((-31*6 )x - (31*5) + (13 * 12)) / 216]# L C M 216.

#y = [(-186x - 155 + 156) / 216]#

#y = (-186x + 1) / 216#

#216y = -186x + 1#

#186x + 216y = 1#