# Shortest distance between a line and a point?

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1. Jul 20, 2017

### Helly123

1. The problem statement, all variables and given/known data
find the shortest distance from (0,0) to the line passing A(2,3) and B(3,5)

2. Relevant equations
$\frac{y-y1}{y2-y1} = \frac{x-x1}{x2-x1}$
y-y1 = m (x-x1)
m1 * m2 = -1 (m1 perpendicular to m2)

3. The attempt at a solution
line passing A and B points
$\frac{y-3}{5-3}$ = $\frac{x-2}{3-2}$
y-3 = 2(x-2)
y = 2x - 1

m1 = 2
m2 = -1/2

the line perpendicular to line 1 and passing (0,0)
y -y1 = m(x-x1)
y-0=-1/2(x-0)
y' = -1/2x'

line 1 and line 2 intersect at (x,y)
y = y', x = x'
2x - 1 = -1/2x'
3/2x = 1
x = 2/3

y' = -1/2'x = -1/2(2/3) = -1/3

x,y = 2/3 , -1/3

distance is from (0,0) to (2/3 , -1/3)
distance = $\sqrt{(2/3-0)^2 + (-1/3-0)^2}$
distance = $\sqrt{4/9 + 1/9}$
distance = $\sqrt{5}/3$

what's wrong? I got wrong answer

2. Jul 20, 2017

### ehild

The red line.

3. Jul 20, 2017

### SammyS

Staff Emeritus
Simply an Algebra mistake. Highlighted in red.

If you graph the line $\ y=2x-1\$, you will notice that the point (2/3, −1/3) does not lie on the line.

Last edited: Jul 20, 2017
4. Jul 20, 2017

### Staff: Mentor

@ehild already pointed out the mistake in the last line above. In addition, the three lines I picked out above are ambiguous, as -1/2x' might be interpreted as $-\frac{1}{2x'}$ by some when you probably meant $-\frac{1}2 x'$. The same goes for 3/2x.

5. Jul 20, 2017

### Helly123

thank you for all the corrections... :)

6. Jul 20, 2017

### SammyS

Staff Emeritus
(1/5) $\sqrt5$