# What are the coordinates of the point on the line 6 x - 5 y = 8 that is closest to the origin?

## Question:

What are the coordinates of the point on the line {eq}6 x - 5 y = 8 {/eq} that is closest to the origin?

## Closest Point on the Line

The closest point on a line relative to a specific coordinate is very important in several applications. One of which is the concept of a tangent of a circle. Knowing the closest point will help solve for the radius of a circle that is tangent on that point on the line.

The closest point on a line that to the origin is the intersection point of the given line and the line passing to the origin which is perpendicular to the given line.

A line is said to be perpendicular to another if their slopes are the negative of the reciprocal of the other. Transforming the given equation into the slope-intercept form, we have

{eq}\begin{align} 6x-5y&=8\\[0.2cm] -5y&=-6x+8\\[0.2cm] y&=\dfrac{6}{5}x-\dfrac{8}{5}\\[0.2cm] y&=mx+b \end{align} {/eq}

Thus the slope of the given line is {eq}\dfrac{6}{5} {/eq}. Therefore, line perpendicular to this has {eq}m=-\dfrac{5}{6} {/eq} and since the reference is the origin, the y-intercept of this line is {eq}b=0 {/eq}, giving us the equation of the line to be:

{eq}y=-\dfrac{5}{6}x {/eq}

Now, we only need to solve where the two lines intersect. We now have two equations given by:

{eq}\begin{align} y&=\dfrac{6}{5}x-\dfrac{8}{5}\\[0.2cm] y&=-\dfrac{5}{6}x \end{align} {/eq}

This means that

{eq}\begin{align} \dfrac{6}{5}x-\dfrac{8}{5}&=-\dfrac{5}{6}x\\[0.2cm] \dfrac{6}{5}x+\dfrac{5}{6}x&=\dfrac{8}{5}\\[0.2cm] x\left(\dfrac{6}{5}+\dfrac{5}{6}\right)&=\dfrac{8}{5}\\[0.2cm] x\left(\dfrac{36}{30}+\dfrac{25}{30}\right)&=\dfrac{8}{5}\\[0.2cm] x\left(\dfrac{36+25}{30}\right)&=\dfrac{8}{5}\\[0.2cm] x\left(\dfrac{61}{30}\right)&=\dfrac{8}{5}\\[0.2cm] x&=\left(\dfrac{8}{5}\right)\left(\dfrac{30}{61}\right)\\[0.2cm] x&=\dfrac{48}{61} \approx 0.79 \end{align} {/eq}

Using the perpendicular line, we have

{eq}\begin{align} y&=-\dfrac{5}{6}x\\[0.2cm] y&=\left(-\dfrac{5}{6}\right) \left(\dfrac{48}{61}\right)\\[0.2cm] y&=\dfrac{40}{61} \approx 0.66 \end{align} {/eq}

Therefore the nearest point on the line to the origin is {eq}(\dfrac{48}{61},\dfrac{40}{61}) {/eq} or about {eq}(0.79,0.66). {/eq}.