#color(blue)("Initial investigation")#
You state that the given equation (an equation has an equals sign in it whilst an expression does not) has the point #(x,y)->(-6-6)# satisfying it. Lets test that
Set point 1 as #P_1->(x,y)=(-6,-6)#
Lets suppose that #P_1# is on the line #y_1=3x_1+c_1#
Then we can substitute #color(red)(-6) # for #color(red)(x_1)# giving:
Using the slope-intercept form #y=mx+c# we have:
#color(green)(y_1=3color(red)(x_1)+13 color(white)("dddd") ->color(white)("dddd")y=3(color(red)(-6))+13)#
#color(green)( color(white)("ddddd.dddddddd.d") ->color(white)("dddd")y_1=-18+13)#
#color(green)( color(white)("dddd.dddddddd.dd") ->color(white)("dddd")y_1=-5)color(red)(larr" Contradiction")#
#color(brown)("Clearly "P_1" does not lie on the line "y_1=3x_1+13)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Investigation a plausible interpretation of the question")#
#color(brown)("Assumption: "P_1" is on a line perpendicular to the given one")#
Known: given that the original equation is of type
#color(white)("dddddddd")y_1=mx_1+c_1#. where #m=3#
Then the line perpendicular to this is of form
#color(white)("dddddddd")y_2=-1/mx_2+c_2#
#color(white)("dddddddd")y_2=-1/3x_2+c_2#
Set #P_1->(x_2,y_2)=(color(red)(-6),color(purple)(-6))#
Then we can substitute #color(red)(-6) # for #color(red)(x_2)# and #color(purple)(-6 )# for #color(purple)(y_2)# giving
#color(green)(y_2=-1/3color(red)(x_2)+c_2 color(white)("dddd") ->color(white)("dddd")color(purple)(-6)=-1/3(color(red)(-6))+c_2)#
#color(white)("dddddddddddddddddd")->color(white)("dddd") color(green)(-6=color(white)("dd")+2color(white)("d.d")+c_2)#
From this we see that #c_2=-8# giving:
#color(white)("dddddddddddddddddd")color(green)(->color(white)("dddd")y_2=-1/3x_2-8)#
Slope intercept form
