**This question aims to find a point that is nearest to the origin and that is lying on the given line $y$ = $5x$ + $3$.**

The **distance formula** is used to calculate the distance between **two sets** of **points** where **( $x_1$ , $y_1$ )** is the first set of points and **( $y_1$ , $y_2$ )** is the other set of points. $d$ is the distance between these points. It is calculated by the formula:

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

The distance of any **point** on the line from the **origin** can be calculated by using the distance formula.

**Expert Answer**

Consider a **point** ($x$ , $y$) on the **line** that is closest to the **origin.** The given line is $y$ = $5x$ + $3$, so the point ($P$) will be written as:

\[P = ( x , y)\]

\[y = 5x + 3\]

By putting the value of y in the point:

\[P = ( x , 5x +3)\]

Assume other **order pair** $(0, 0)$.

By using **distance formula:**

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

By putting the set of **ordered pairs** ( $x$ , $5x$ + $3$ ) and ( $0$ , $0$) in the distance formula:

\[d = \sqrt{( x – 0 )^2 + ( 5x + 3 – 0 )^2}\]

\[d = \sqrt{x^2 + (25 x^2 + 30 x + 9) }\]

\[d = \sqrt{ 26 x^2 + 30 x + 9}\]

By putting $d’$** =** $0$ and using **chain rule,** the **derivative** will be:

\[d’ = \frac{1}{2} (26 x^2 + 30 x + 9)^ {\frac{-1}{2}} \times \frac{d}{dx} (26 x^2 + 30 x + 9)\]

\[= \frac{1}{2 \sqrt{26 x^2 + 30 x + 9}} \times 52 x + 30 + 0\]

\[d’ = \frac{52 x + 30}{2 \sqrt{26 x^2 + 30 x + 9}}\]

By putting $d’$ = $0$, we get:

\[0 = \frac{52 x + 30}{2 \sqrt{26 x^2 + 30 x + 9}}\]

By multiplying the **denominator** with the number on left hand side:

\[0 \times 2 \sqrt{26 x^2 + 30 x + 9} = 52 x + 30\]

\[0 = 52 x + 30\]

\[-30 = 52 x\]

\[\frac{-30}{52} = x\]

\[x = \frac{-15}{26}\]

The graph above is showing the point $x$ = $\frac{-15}{26}$, **plotted** on the **line** $y$ = $5x$ + $3$.

## Numerical Results

Hence, the **point lying** on the line and **nearest** to the **origin** is $\frac{-15}{26}$.

## Example

The **distance** of two set of points ($1$ , $2$) and ($3$ , $4$) is calculated by:

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

\[d = \sqrt{(3 – 1)^2 + (4 – 2)^2}\]

\[d = \sqrt{4 + 4}\]

\[d = \sqrt{8}\]

\[d = 2 \sqrt{2}\]

**The distance between two points is $2 \sqrt{2}$.**

*Images/Mathematical drawings are created in Geogebra.*