Answer:
[tex]distance=\frac{27}{53}\ units\approx0.509\ units[/tex]
Step-by-step explanation:
You need to use the following formula for calculate the distance between a point and a line:
[tex]d=\frac{ |Ax_1+By_1+C|}{\sqrt{A^2+B^2}}[/tex]
Rewrite the equation of the given line in the following form:
[tex]Ax+By+C=0[/tex]
Then:
[tex]\frac{45}{28}x+y+\frac{25}{28}=0[/tex]
You can identify that:
[tex]A=\frac{45}{28}\\\\B=1\\\\C=\frac{25}{28}[/tex]
Given the point:
[tex](\frac{1}{5},-\frac{1}{4})[/tex]
You can identify that:
[tex]x_1=\frac{1}{5}\\\\y_1=-\frac{1}{4}[/tex]
Therefore, you can substitute values into the formula:
[tex]d=\frac{ |(\frac{45}{28})(\frac{1}{5})+(1)(-\frac{1}{4})+\frac{25}{28})|}{\sqrt{(\frac{45}{28})^2+(1)^2}}=\frac{27}{53}\ units\approx0.509\ units[/tex]
