Point $P(2,5)$ to the line $y=2x-3$

A point $P(2,5)$ to the line by $y=2x-3$

The goal is to calculate the distance between the point and the line.

The formula for the distance is $\frac{||d\times \overrightarrow{P_{0}P}||}{\Vert d\Vert }$

Assume that, $x=t$

Then the equation $y=2x-3$becomes,

$y=2t-3$for some parameter $t$

Now the line equation in vector parametrization is $r\left(t\right)=(t,2t-3)$

Let the point $P$is $P(2,5)$

The point $P_0$ on the line equation $r\left(t\right)=(t,2t-3)$ is $(0,-3)$

The direction vector $\overrightarrow{P_{0}P}=(2-0,5-(-3\left)\right)$

Then, $\overrightarrow{P_{0}P}=(2,8)$

Take the direction vector of the equation $r\left(t\right)=(t,2t-3)$is $d=(1,2)$

Substitute the values $d=(1,2)$ and $\overrightarrow{P_{0}P}=(2,8)$in the formula $\frac{\Vert d\times \overrightarrow{P_{0}P}.}{\Vert d\Vert }$

Then the distance $=\frac{\Vert (1,2)\times (2,8)\Vert }{\Vert (1,2)\Vert }$

Distance $=\frac{|2·2-1·8|}{\sqrt{1^2+2^2}}$

Thus,

Multiply and divide by $\sqrt{5}$

Distance $=\frac{4}{\sqrt{5}}·\frac{\sqrt{5}}{\sqrt{5}}$

Distance $=\frac{4\sqrt{5}}{5}$

The distance from the point $P(2,5)$ to the line $y=2x-3$ is $\frac{4\sqrt{5}}{5}$ units.

Therefore, the answer is $\frac{4\sqrt{5}}{5}$ units