Question

(1) Let E⊆R² be given by E={(x,y)∈R²:(x,y) is equidistant from (1,4) and (7,2)}. Let S⊆R² be given by S={(3+t,3t)∈R2:0≤t≤1}. Prove that S⊆E. Also, show that E∈/S.

Answer 1

To prove that S⊆E, we show that every point in S is also in E by comparing the distances between the points in S and the given points. To demonstrate that E∉S, we find a point in E that doesn't fit the required form of S.

Step-by-step explanation:

To prove that S⊆E, we need to show that every point in S is also in E. Let's take a point (3+t,3t) from S. Now, we need to show that this point is equidistant from (1,4) and (7,2).

Let's find the distance between the point (3+t,3t) and (1,4):

d₁ = sqrt((3+t-1)² + (3t-4)²) = sqrt(t²+10t+16)

Now, let's find the distance between the point (3+t,3t) and (7,2):

d₂ = sqrt((3+t-7)² + (3t-2)²) = sqrt(t²-8t+13)

We can see that d₁ = d₂, which means the point (3+t,3t) is equidistant from (1,4) and (7,2). Therefore, S⊆E.

To show that E∉S, we need to find a point in E that is not in S. Let's consider the point (4,6). To be in S, the x-coordinate of the point should be 3+t, but in this case, the x-coordinate is 4, which is not of the required form. Therefore, E∉S.