#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)#