# Two-way bounding problem 3

Here are the set definitions again for reference:

• $$A = \{(x,y) \in \mathbb{R}^2 : y = 3x + 7\}$$
• $$B = \{\lambda(-2,1) + (1-\lambda)(1,10) : \lambda \in \mathbb{R}\}$$

### Solution

Lemma: $$\lambda(-2,1) + (1-\lambda)(1,10) = (-2\lambda + (1 - \lambda), \lambda + (10 - 10\lambda)) = (1 - 3\lambda, 10 - 9\lambda)$$

$$A \subseteq B$$

Let (x,y) be an element of A. Then $$y = 3x + 7$$ by the definition of A.

Consider $$\lambda = \frac{1-x}{3}$$.

Then, using our lemma above, we can calculate:

$$\lambda(-2,1) + (1-\lambda)(1,10) = (1 - 3\lambda, 10 - 9\lambda) = (1 - 3\frac{1-x}{3}, 10 - 9\frac{1-x}{3}) = (1 - (1-x), 10 - 3(1-x)) = (x, 3x+7) = (x,y)$$

So we've shown that (x,y) is an element of B.

Since every element of A is also an element of B, $$A \subseteq B$$.

$$B \subseteq A$$

Let (x,y) be an element of B. By the definition of B, we know that $$(x,y) = \lambda(-2,1) + (1-\lambda)(1,10)$$ for some real number $$\lambda$$.

So, using our lemma above, $$(x,y) = (1 - 3\lambda, 10 - 9\lambda)$$.

Then $$3x+7 = 3(1 - 3\lambda) + 7 = 3 - 9\lambda + 7 = 10 - 9\lambda = y$$. So 3x+7 = y, which means that (x,y) is an element of A.

Since every element of B is also an element of A, $$B \subseteq A$$.

Since $$A \subseteq B$$ and $$B \subseteq A$$, A=B, which is what we needed to show.