Am I right or wrong in this absolute value?

Question

I had a question to find $p$ which was a co-ordinate in a question where the other coordinates were given and area of triangle was given.

After simplifying the last equation came as: $$20 = \vert 13p - 20\vert .$$ Now $20 = |20| = |-20|$. So $p = 0$ and $p = 40/13$ should be the two answers, but my Maths teacher who only wants students to use her methods and cuts other methods cut $40/13$ and said that only 0 was correct.

I am sure I was right, please tell me whether I am wrong or right.

Edit: I will post the question as soon as I get it. I dont have my question paper now.

Edit: Question reads, "Find the value of $p$ for which the area of triangle with vertices $A(p, 2p), B (-2,6) and C (3,1) is 10 sq. units$.

But shouldn't there be infinite values of p? I mean if we take BC as base then we can find many points A with height of triangle same. Please tell me, thank you.

Please answer with whole question solved with proof that I am right so that I can print it. I can print it, right? Or do I need permission? And yes, if you can provide graph in the answer, that will be awesome as well.

Edit: The issue has been resolved. Thank You.

Answer 1

You are right indeed.

• for $13p - 20>0 \implies 20 = 13p - 20 \implies p=\frac{40}{13}$
• for $13p - 20<0 \implies 20 = -13p + 20 \implies p=0$

Maybe there is some other motivation to exclude $p=\frac{40}{13}$ depending on the context of the problem.

Answer 2

The exclusion of the other solution might be actually motivated by some other context of your exercise like constrains on $p$ or some previous questions which you did mentioned here. But mathematically both solution are correct as well.

However, you are right but here is one way to get both solution at once solution

$$|13p-20|=20 \Longleftrightarrow (13p-20=20 ~~or ~~~~13p-20=-20) $$

that is $p =0~~or~~~~p =\frac{40}{13}$

Edit (After OP added missing information):

It is obvious to see geometrically that there must two triangles which are symmetric to each other with respect to the line (BC) therefore you are totally right