Find the point on the line yequalsx that is closest to the point left parenthesis 5 comma 1 right parenthesis .
​[Hint​:  Express the distance d from the point to the line as a function of​ x, and then find the minimum value of ​[d(x)]squared​.]

To solve this problem, we need to find the point on the line \( y = x \) that is closest to the point \( (5, 1) \). We'll do this by minimizing the distance between a generic point on the line \( y = x \) and the point \( (5, 1) \).

### Step 1: Express the Distance \( d \) as a Function of \( x \)

Let \( P(x, y) \) be any point on the line \( y = x \). The distance \( d \) between \( P(x, x) \) and \( (5, 1) \) is given by the distance formula:
\[
d = \sqrt{(x - 5)^2 + (x - 1)^2}
\]
Since we are minimizing the square of the distance, we can simplify our work by considering the square of the distance \( d^2 \) (denoted as \( f(x) \)) instead of \( d \):
\[
f(x) = (x - 5)^2 + (x - 1)^2
\]

### Step 2: Simplify the Expression

Expand and simplify \( f(x) \):
\[
f(x) = (x - 5)^2 + (x - 1)^2
\]
\[
f(x) = (x^2 - 10x + 25) + (x^2 - 2x + 1)
\]
\[
f(x) = x^2 - 10x + 25 + x^2 - 2x + 1
\]
\[
f(x) = 2x^2 - 12x + 26
\]

### Step 3: Find the Minimum Value of \( f(x) \)

To find the minimum value of \( f(x) \), take the derivative with respect to \( x \) and set it equal to zero:
\[
f'(x) = 4x - 12
\]
Set \( f'(x) = 0 \) to find the critical point:
\[
4x - 12 = 0
\]
\[
x = \frac{12}{4} = 3
\]

### Step 4: Find the Corresponding \( y \) Value

Since the point lies on the line \( y = x \), the \( y \) value corresponding to \( x = 3 \) is also \( y = 3 \).

### Step 5: Conclusion

The point on the line \( y = x \) that is closest to \( (5, 1) \) is \( (3, 3) \).

Would you like further details or have any questions? Here are some related questions:

1. How can we confirm that \( x = 3 \) is indeed a minimum point?
2. What is the minimum distance between the point \( (5, 1) \) and the line \( y = x \)?
3. How would the approach change if the line had a different slope?
4. How does this problem relate to the concept of perpendicular distance?
5. What if the point were on the line \( y = -x \) instead?

**Tip:** When finding the closest point on a line to a given point, remember that the line segment connecting them is perpendicular to the line.

Find the point on the line y = x that is closest to the point (5, 1). [Hint: Express the distance d from the point to the line as a function of x, and then find the minimum value of [d(x)] squared.]

This problem demonstrates how to find the point on the line y = x that is closest to (5, 1) by minimizing the distance using the distance formula and derivative calculus. Step-by-step solution provided for high school students learning minimization techniques and linear functions.

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