The aim of this problem is to find a **point** that is **nearest **to the **origin**. We are given a linear equation which is only a **straight line** in the xy-plane. The **nearest** point from the origin will be the **vertical** distance from the origin to that line. For this, we need to be aware of the **distance formula** between two points and the **derivation**.

The **nearest distance** of a point to a line will be the **smallest vertical** distance from that point to any random point on the straight line. As concerned above, it is the** perpendicular** distance of the point to that line.

To solve this problem, we will have to figure out an **equation** of the perpendicular from (0,0) on y = 4x + 3. This equation is actually the **slope intercept form** i.e. y = mx + c.

## Expert Answer

Let’s assume $P$ to be the **point** that is on the line $y = 4x+3$ and closest to the **origin**.

Suppose the $x$-**coordinate** of $P$ is $x$ and $y$-**coordinate** is $4x+3$. So the point is $(x, 4x+3)$.

We have to find the **distance** of point $P (x, 4x+3)$ to the origin $(0,0)$.

**Distance formula** between two points $(a, b)$ and $(c, d)$ is given as:

\[D=\sqrt{(a + b)^2+(c + d)^2 }\]

Solving it for $(0,0)$ and $(x, 4x+3)$:

\[D=\sqrt{(x-0)^2+(4x+3 -0)^2 }\]

\[=\sqrt{x^2+(4x+3)^2 }\]

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

Now let:

\[f(x)=\sqrt{x^2 + (4x+3)^2 }\]

We have to find the $x$ that makes $f(x)$ minimum by implementing a **derivation.**

If we minimize $x^2 + (4x+3)^2$, it will automatically **minimize** the $\sqrt{x^2 + (4x+3)^2 }$ so assuming $x^2 + (4x+3)^2$ to be $g(x)$ and minimizing it.

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

\[g(x)=x^2+16x^2+9+24x\]

\[g(x)=17x^2+24x+9\]

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

\[g'(x)=34x + 24\]

\[0 = 34x + 24\]

$x$ comes out to be:

\[x=\dfrac{-12}{17}\]

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

\[P=(x, 4x+ 3)\]

\[=(\dfrac{-12}{17} , 2(\dfrac{-12}{17})+ 3)\]

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

\[P=(\dfrac{-12}{17},\dfrac{27}{17})\]

## Numerical Result

$(\dfrac{-12}{17},\dfrac{27}{17})$ is the **point** on the line $y = 4x+3$ that is **closest** to the **origin**.

## Example

Find a point on a **straight** **line** $y = 4x + 1$ that is **nearest** to the origin.

Let’s assume $P$ to be the point $(x, 4x+1)$.

We have to find the **smallest distance** of point $P (x, 4x+1)$ from the origin $(0,0)$.

\[D=\sqrt{x^2 + (4x+1)^2 }\]

Now let,

\[f(x)=\sqrt{x^2 +(4x+1)^2 }\]

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

Let’s assume,

\[g(x)=x^2 + (4x+1)^2 \]

\[g(x)= x^2 + 16x^2+ 1 + 8x \]

\[g(x) = 17x^2 +8x + 1\]

Taking **derivative** of $g(x)$ and put it equals to $0$.

\[g'(x) = 34x + 8\]

\[0 = 34x + 8 \]

$x$ comes out to be:

\[x = \dfrac{-4}{17} \]

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

\[P=(x, 4x+ 1) \]

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

\[P=( \dfrac{-4}{17} , \dfrac{1}{17})\]