Respuesta :
Given:
Point is [tex]A\left(-\dfrac{1}{4},5\right)[/tex].
Line is [tex]-x+2y=14[/tex].
To find:
Distance from point A to given line.
Solution:
The distance of point [tex](x_0,y_0)[/tex] from line [tex]ax+by+c=0[/tex] is
[tex]d=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}[/tex]
Using the above formula, the distance from [tex]A\left(-\dfrac{1}{4},5\right)[/tex] to the line [tex]-x+2y-14=0[/tex] is
[tex]d=\dfrac{|-(-\dfrac{1}{4})+2(5)-14|}{\sqrt{(-1)^2+(2)^2}}[/tex]
[tex]d=\dfrac{0.25+10-14|}{\sqrt{1+4}}[/tex]
[tex]d=\dfrac{|-3.75|}{\sqrt{5}}[/tex]
[tex]d=\dfrac{3.75}{\sqrt{5}}[/tex]
[tex]d=1.677[/tex]
[tex]d\approx 1.7[/tex]
Therefore, the required distance is 1.7 units.
