This problem aims to find a **point** that is closest to the origin. A **linear equation** is given, which is just a simple line in the xy-plane. The closest point from the origin will be the **vertical distance** from the origin to that line. For this, we need to be familiar with the **distance formula** between two points and the **derivatives**.

The distance from a line to a point is the **smallest distance** from a point to any arbitrary point on a straight line. As discussed above, it is the **perpendicular** distance of the point to that line.

We need to figure out an equation of the **perpendicular** from (0,0) on y = 2x + 3. This equation is of the **slope intercept** form i.e. y = mx + c.

## Expert Answer

Let’s **assume** $P$ to be the point that is on the line $y = 2x+3$ and closest to the origin.

Suppose the $x$-**coordinate** of $P$ is $x$ and $y$-**coordinate** is $2x+3$. So the point is $(x, 2x+3)$.

We have to find the **distance** of point $P (x, 2x+3)$ to the origin $(0,0)$.

**Distance** **f****ormula** between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given as:

\[D=\sqrt{(x_1 + x_2)^2+(y_1 + y_2)^2 }\]

Solving it for $(0,0)$ and $(x, 2x+3)$:

\[D=\sqrt{(x-0)^2+(2x+3 -0)^2 }\]

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

We have to **minimize** the $x$ to find the **minimal** distance from point $P$ to the origin.

Now let:

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

We have to find the $x$ that makes $f(x)$ smallest by usual **derivative** process.

If we **minimize** $x^2 + (2x+3)^2$, it will automatically **minimize** the $\sqrt{x^2 + (2x+3)^2 }$ so assuming $x^2 + (2x+3)^2$ to be $g(x)$ and minimizing it.

\[g(x)=x^2 + (2x+3)^2\]

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

\[g(x)=5x^2+12x+9\]

To find the minimum lets take the **derivative** of $g(x)$ and put it equals to $0$.

\[g'(x)=10x + 12\]

\[0 = 10x + 12\]

$x$ comes out to be:

\[x=\dfrac{-6}{5}\]

Now put $x$ into the **point** $P$.

\[P=(x, 2x+ 3)\]

\[=(\dfrac{-6}{5} , 2(\dfrac{-6}{5})+ 3)\]

**Point** $P$ comes out to be:

\[P=(\dfrac{-6}{5},\dfrac{3}{5})\]

## Numerical Result

$(\dfrac{-6}{5},\dfrac{3}{5})$ is the **point** on the line $y = 2x+3$ that is **closest** to the **origin.**

## Example

Find the **point **that is closest to the origin and lie on the line $y = 4x + 5$.

Let’s assume $P$ to be the point is $(x, 4x+5)$.

We have to find the **distance** of point $P (x, 4x+5)$ to the **origin** $(0,0)$.

\[D=\sqrt{x^2 + (4x+5)^2 }\]

Now let:

\[f(x)=\sqrt{x^2 +(4x+5)^2 }\]

We have to find the $x$ that makes $f(x)$ **smallest** by usual derivative process.

Let’s assume,

\[g(x) = x^2 + (4x+5)^2 \]

\[g(x) = x^2 + 16x^2+ 25 + 40x \]

\[g(x) = 17x^2 +40x + 25\]

To find the **minimum** lets take the **derivative** of $g(x)$ and put it equals to $0$.

\[g'(x) = 34x + 40\]

\[0 = 34x + 40 \]

$x$ comes out to be:

\[x = \dfrac{-20}{17} \]

Now put $x$ into the point $P$.

\[P = (x, 4x+ 5) \]

**Point** $P$ comes out to be:

\[P = ( \dfrac{-20}{17} , \dfrac{5}{17})\]