Find the distance between point ( 1/5 , − 1/4 ) and line y=− 45/28 x− 25/28

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]