To find the point on the line $y = 5x + 4$ that is closest to the origin, we can follow these steps:

1. Set Up the Distance Formula: The distance from a point $(x, y)$ on the line to the origin $(0, 0)$ is given by: $d = \sqrt{x^2 + y^2}$ Since $y = 5x + 4$, we can substitute $y$ in terms of $x$: $d = \sqrt{x^2 + (5x + 4)^2}$

2. Minimize the Distance: To find the point closest to the origin, we need to minimize $d^2$ (which is equivalent to minimizing $d$ since the square root is a monotonically increasing function): $d^2 = x^2 + (5x + 4)^2$ Expanding this: $d^2 = x^2 + (25x^2 + 40x + 16)$ $d^2 = 26x^2 + 40x + 16$

3. Take the Derivative and Set It to Zero: Now, differentiate $d^2$ with respect to $x$ and set the derivative equal to zero to find the minimum. $\frac{d(d^2)}{dx} = 52x + 40 = 0$ Solving for $x$: $x = -\frac{40}{52} = -\frac{10}{13}$

4. Find the Corresponding $y$-Value: Substitute $x = -\frac{10}{13}$ back into the line equation $y = 5x + 4$: $y = 5\left(-\frac{10}{13}\right) + 4 = -\frac{50}{13} + \frac{52}{13} = \frac{2}{13}$

5. Solution: The point on the line $y = 5x + 4$ that is closest to the origin is: $\left( -\frac{10}{13}, \frac{2}{13} \right)$

Would you like more details on any step?

Related Questions

1. How do you derive the distance formula for a point and the origin?
2. Why do we use the derivative to find minimum distances?
3. What would change if the line equation had a different slope?
4. How can we generalize this method for any line equation?
5. Why does minimizing $d^2$ instead of $d$ work?

Tip

In optimization problems involving distance, minimizing $d^2$ is often simpler than working directly with $d$, as it avoids the square root.