#color(red)("Assumption: This question is based on a straight line")#
The standard form of this equation type is:#" "y=mx+c#
Where #m# is the gradient and #c# is the y intercept
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To determine the gradient")#
The x-intercept is at #y=0#
Let point #P_1# be the x-intercept
Then #P_1->(x_1,y_1)=(2,0)#
We are given a point A
Let point #P_A->(x_A,y_A)=(-6,8)#
#color(brown)("It is not good practice to change the lettering in a question.")# #color(brown)("It makes a lot of work for the person marking your solution")#
The gradient is #("change in y")/("change in x")=m#
#=>m->P_A-P_1-> ("change in y")/("change in x")= (y_A-y_1)/(x_A-x_1 #
#=> m=(8-0)/(-6-2)=8/(-8)=-1#
#color(green)("So the equation is now "y=-x+c)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(white)(.)#
#color(blue)("Determine the value of "c)#
The line passes through #P_A->(x,y)=(-6,8)#
So by substitution we have
#8=-(-6)+c#
#=>8=+6+c#
Subtract 6 from both sides
#=>c=2#
#color(green)("So the equation is now "y=-x+2)#