Distance between a line and a point?

Question

What is the shortest distance from (0 ,0) to the line 5x + 12y - 39 = 0. I am so lost.

Answer 1

Using a different method that requires Calculus skills:

Minimum Distance = 3

The distance = sqrt(x^2+y^2) = sqrt(y^2+(7.8-2.4y)^2)
The distance^2 = y^2+60.84-37.44y+5.76y^2

When the distance is minimized, so is the distance squared...it's easier to determine.

The minimum occurs when the derivative = 0

Distance^2 derivative = -37.44+13.52y = 0
Therefore, y = 37.44/13.52 = 2.77
and x = 7.8-2.4*(2.77) = 1.154

Thus the distance = sqrt(2.77^2+1.154^2) = 3

Answer 2

First put the equation into the form y = mx + b

5x + 12y - 39 = 0
12y = -5x + 39
y = -5x/12 + 39/12

The slope of the given line is -5/12

The shortest distance is the line going through (0, 0) that is perpendicular to the given line. The slope of this perpendicular line is +12/5 - the negative reciprocal of the given line's slope

Since the new line goes through (0, 0), it's y-intercept is 0, so the new equation is

y = 12x/5

We need to find the point of intersection of these two lines, so solve the system

y - 12x/5 = 0
y + 5x/12 = 39/12

Subtract the second equation from the first

-12x/5 - 5x/12 = - 39/12

Multiply both sides by the LCD of 60

-144x - 25x = - 195
-169x = -195
x = 195/169 = 15/13

y = 12x/5
y = (12/5)(15/13) = 36/13

The lines intersect at (15/13, 36/13), and the distance from this point to (0, 0) is

sqrt[(15/13)^2 + (36/13)^2]
sqrt[(225 + 1296)/13^2]
sqrt(1521/13^2]
sqrt(39^2/13^2)
39/13
3

Distance is 3

Answer 3

In general if you have a line
ax+by+c=0 and a point (xo,yo) the distance of the point to the line is given by
d = abs val (axo+byo+c)/sqrt(a^2+b^2)
In your case the point is (0,0) so d=abs val (5*0+12*0-39)/sqrt(5^2+12^2) = 39/13 =3

Answer 4

5x + 12y = 39 has slope = -5/12
the perpendicular line from (0,0) to the line has slope 12/5
so
5x + 12y = 39
and 5y = 12x are the two lines
solving
60y = 144x
25x + 60y = 195
169x = 195
x = 195/169 = 1.154
and y = 12x/5 = 2.769
is the point they meet
so distance from (0,0) = sqrt (1.154^2 + 2.769^2) = 65.318