Melody

I expect you are missing brackets.

Even if you are not you should put them in so people are sure what your question is.

I know people here often just interpret the question but maths is precise and you should learn to ask your question in a truly unambiguous way.

y=-3x+8

make x the subject then swap the x and y over.   x>2

This is what your graph will look like.

Thanks Pangolin14  and Yugoslav     

I tackled this in a different way.

I also used co-ordinate geometry.  As shown on my pic.

I set the centre of the incircle (since that is what I is) as the origin.

B(-4,-3)  and  C((4,-3)    

I need to find the height of the triangle 

Let the equation of the line AB be     y= -mx+b  where m and b are both positive.

The perpendicular distance of the origin to line AB is the radius, 3.

So I will use the perpendicular distance formula

y= -mx+b

mx + y -b = 0        (0,0)

\(3=\frac{|0+0-b|}{\sqrt{m^2+1}}\\ 3=\frac{b}{\sqrt{m^2+1}}\\ b=3\sqrt{m^2+1} \)

So we have

\(y=-mx+3\sqrt{m^2+1}\\\)

This line passes through the point (4,-3) so substitute

\(-3=-4m+3\sqrt{m^2+1}\\ -3+4m=3\sqrt{m^2+1}\\ 16m^2-24m+9=9(m^2+1)\\ 16m^2-24m+9=9m^2+9\\ 7m^2-24m=0\\ 7m-24=0\\ m=\frac{24}{7}\)

\(b=3\sqrt{m^2+1}\\ b=3\sqrt{(\frac{24}{7})^2+1}\\ b=3*\frac{25}{7}=\frac{75}{7}\\ \text{height of triangle }= \frac{75}{7}+3 = \frac{96}{7}\)

\(\text{Area of triangle ABC = }\frac{1}{2}*8*\frac{96}{7}\\ \text{Area of triangle ABC = }54\frac{6}{7}\;units^2\)

W+3=L

W(W+3)=70

Solve it as a quadratic.

Pangolin,

You have not explained how you got the point A.

Remember, no trig is allowed.

Once again JoLink you have stolen another person's work.

This is a copy of CoolStuffYT's answer on July 20.

Thank you Pangolin14    

See CPhill's answer

It is good that you like it Chris, it is your answer.