25. Find the point on the line y = 2x + 3 that is closest to the
origin.

To find the point on the line \( y = 2x + 3 \) that is closest to the origin, we need to minimize the distance from a point \( (x, y) \) on the line to the origin \( (0, 0) \). The distance \( d \) from the origin to a point \( (x, y) \) is given by:

\[
d = \sqrt{x^2 + y^2}
\]

Since \( y = 2x + 3 \), we can substitute for \( y \) in the distance formula:

\[
d = \sqrt{x^2 + (2x + 3)^2}
\]

To minimize \( d \), it suffices to minimize \( d^2 = x^2 + (2x + 3)^2 \) because the square root is a monotonically increasing function.

Expand \( (2x + 3)^2 \):

\[
(2x + 3)^2 = 4x^2 + 12x + 9
\]

Thus, \( d^2 \) becomes:

\[
d^2 = x^2 + 4x^2 + 12x + 9 = 5x^2 + 12x + 9
\]

To find the minimum, take the derivative of \( d^2 \) with respect to \( x \) and set it to zero:

\[
\frac{d}{dx}(5x^2 + 12x + 9) = 10x + 12
\]

Set the derivative equal to zero:

\[
10x + 12 = 0
\]

Solve for \( x \):

\[
10x = -12 \\
x = -\frac{6}{5}
\]

Substitute \( x = -\frac{6}{5} \) back into the line equation to find \( y \):

\[
y = 2\left(-\frac{6}{5}\right) + 3 = -\frac{12}{5} + \frac{15}{5} = \frac{3}{5}
\]

Therefore, the point on the line \( y = 2x + 3 \) that is closest to the origin is:

\[
\left(-\frac{6}{5}, \frac{3}{5}\right)
\]

Find the point on the line y = 2x + 3 that is closest to the origin.

Learn how to determine the point on the line y = 2x + 3 that is closest to the origin using the distance formula and optimization techniques. Includes step-by-step explanations of algebraic methods and calculus concepts. Ideal for high school students in Grades 10-12.

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