 # Find the point on the line y=5x+3 that is closest to the origin. 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.

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:

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

$0 = 52 x + 30$

$-30 = 52 x$

$\frac{-30}{52} = x$

$x = \frac{-15}{26}$ Figure 1

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.