Given is the convex polyhedron $P:= \left\{\begin{pmatrix} x\\ y \end{pmatrix} \in \mathbb{R}^2 \mid 5x-2y \leq 7, x \geq 0, y \geq 0\right\}$. Show that $A = \begin{pmatrix} 3\\ 4 \end{pmatrix}$ is not an extreme point in $P$ and that the point $B = \begin{pmatrix} 1.4\\ 0 \end{pmatrix}$ is an extreme point.
I have a definition about an extreme point from my book: Point $x$ of a convex set $S \subseteq R^n$ is called an extreme point of $S$ if there is no $u \in R^n$, $u \neq 0_n$ such that there is an $\varepsilon > 0$ with $x+\varepsilon u \in S$ and $x-\varepsilon u \in S$
So I'm not sure if the following counts as "Show" but graphically, $P$ is a triangle where $A$ is an interior point of $5x-2y = 7$ and if you choose $u=(2,5)$ and $\varepsilon$ small enough such that you have that $x + \varepsilon u$ and $x - \varepsilon u$ on $5x-2y=7$. Thus $A$ cannot be an extreme point in $P$.
And $B$ is an extreme point in $P$ because it is a vertex and there is no section in $P$ that contains $B$ in the interior.
I'm not sure if it's correct like that?
Suppose on the contrary that such $u \ne 0$ exists for point $B$, then
$$5(1.4+ \epsilon u_1)-2(0 +\epsilon u_2) \le 7\tag{1}$$ $$5(1.4- \epsilon u_1)-2(0 -\epsilon u_2) \le 7\tag{2}$$ $$\epsilon u_2 \ge 0\tag{3}$$
$$-\epsilon u_2 \ge 0\tag{4}$$
From $(3)$ and $(4)$, we conclude that $u_2=0.$
Hence now, we have $$7+5\epsilon u_1 \le 7 \iff \epsilon u_1 \le 0$$
$$7-5\epsilon u_1 \le 7 \iff \epsilon u_1 \ge 0$$
whcih again enable us to conclude that $u_1 = 0$ which is a contradiction.
Remark for your attempt for point $A$:
It would be better if you can state clearly how small do you want to choose your $\epsilon$ to be.