Answer:
The distance from point A to given line is: 2√5 or 4.472 units
Step-by-step explanation:
Given equation of line is:
[tex]-x+2y= 14[/tex]
And the point is:
[tex]A = (x_1,y_1) = (-14,5)[/tex]
The standard form of equation of line is:
[tex]Ax+By+C = 0[/tex]
Converting the given equation into standard form
[tex]-x+2y-14=0[/tex]
The distance of a point (x1,y1) from a line is given by the formula:
[tex]d = \frac{|Ax_1+By_1+C}{\sqrt{A^2+B^2}}[/tex]
In the given details,
A = -1, B = 2, C = -14
x1 = -14, y1 = 5
Putting the values in the formula
[tex]d = \frac{|(-1)(-14)+(2)(5)-14|}{\sqrt{(-1)^2+(2)^2}}\\d = \frac{14+10-14}{\sqrt{1+4}}\\d = \frac{10}{\sqrt{5}}\\d = \frac{5*2}{\sqrt{5}}\\d = \frac{\sqrt{5}*\sqrt{5}*2}{\sqrt{5}}\\d = 2\sqrt{5}\\d = 4.472\ units[/tex]
Hence,
The distance from point A to given line is: 2√5 or 4.472 units
