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Show that the point (7, 5) is equidistant from the points (2, 4) and (6, 10).

Sagot :

Sure! Let's show that the point [tex]\((7, 5)\)[/tex] is equidistant from the points [tex]\((2, 4)\)[/tex] and [tex]\((6, 10)\)[/tex]. We'll calculate the distances from [tex]\((7, 5)\)[/tex] to each of these two points and then compare them.

### 1. Distance Formula

The distance [tex]\(d\)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] in the coordinate plane is given by the formula:

[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]

### 2. Distance from [tex]\((7, 5)\)[/tex] to [tex]\((2, 4)\)[/tex]

Let's calculate the distance between the points [tex]\((7, 5)\)[/tex] (Point A) and [tex]\((2, 4)\)[/tex] (Point B).

#### Step-by-Step Calculation:

- [tex]\( x_1 = 7 \)[/tex]
- [tex]\( y_1 = 5 \)[/tex]
- [tex]\( x_2 = 2 \)[/tex]
- [tex]\( y_2 = 4 \)[/tex]

Using the distance formula:

[tex]\[ d_{AB} = \sqrt{(2 - 7)^2 + (4 - 5)^2} \][/tex]

Calculate the differences:
[tex]\[ 2 - 7 = -5 \][/tex]
[tex]\[ 4 - 5 = -1 \][/tex]

Square the differences:
[tex]\[ (-5)^2 = 25 \][/tex]
[tex]\[ (-1)^2 = 1 \][/tex]

Sum the squares:
[tex]\[ 25 + 1 = 26 \][/tex]

Take the square root:
[tex]\[ d_{AB} = \sqrt{26} \approx 5.099 \][/tex]

Thus, the distance from [tex]\((7, 5)\)[/tex] to [tex]\((2, 4)\)[/tex] is approximately [tex]\(5.099\)[/tex].

### 3. Distance from [tex]\((7, 5)\)[/tex] to [tex]\((6, 10)\)[/tex]

Now let's calculate the distance between the points [tex]\((7, 5)\)[/tex] (Point A) and [tex]\((6, 10)\)[/tex] (Point C).

#### Step-by-Step Calculation:

- [tex]\( x_1 = 7 \)[/tex]
- [tex]\( y_1 = 5 \)[/tex]
- [tex]\( x_2 = 6 \)[/tex]
- [tex]\( y_2 = 10 \)[/tex]

Using the distance formula:

[tex]\[ d_{AC} = \sqrt{(6 - 7)^2 + (10 - 5)^2} \][/tex]

Calculate the differences:
[tex]\[ 6 - 7 = -1 \][/tex]
[tex]\[ 10 - 5 = 5 \][/tex]

Square the differences:
[tex]\[ (-1)^2 = 1 \][/tex]
[tex]\[ 5^2 = 25 \][/tex]

Sum the squares:
[tex]\[ 1 + 25 = 26 \][/tex]

Take the square root:
[tex]\[ d_{AC} = \sqrt{26} \approx 5.099 \][/tex]

Thus, the distance from [tex]\((7, 5)\)[/tex] to [tex]\((6, 10)\)[/tex] is also approximately [tex]\(5.099\)[/tex].

### 4. Conclusion

Since the distances [tex]\( d_{AB} \)[/tex] and [tex]\( d_{AC} \)[/tex] are both approximately [tex]\(5.099\)[/tex], we can conclude that the point [tex]\((7, 5)\)[/tex] is equidistant from the points [tex]\((2, 4)\)[/tex] and [tex]\((6, 10)\)[/tex].

Thus, the point [tex]\((7, 5)\)[/tex] is indeed equidistant from the points [tex]\((2, 4)\)[/tex] and [tex]\((6, 10)\)[/tex].