dtermine the distance:
x=7y-3
(1,-1)

Question

anonymous · Answer

is this question complete?

dumbcow · Answer

oh the distance from the point to the line
Find perpendicular line that goes through (1,-1)
then find point where 2 lines intersect.
use distance formula to find distance between this point and (1,-1)

anonymous · Answer

is it 11/(5sqrt(2)) ?

dumbcow · Answer

well that would give you the shortest distance,

anonymous · Answer

http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html

dumbcow · Answer

i get
$$d = \frac{\sqrt{11^{2}+77^{2}}}{50}\approx 1.55$$

akshay_budhkar · Answer

well the answer is 11/sqrt(50)... The formula is [x1-7y1+3]/sqrt[1+7^2]..... where x1,y1 is the point,, this is for this particular case.

dumbcow · Answer

cool i didn't know about the formula
either way its the same answer :)

anonymous · Answer

x=7y-3
y= x/7 +3/7
then the normal of the line is y=(-7)x+c
as, this  normal goes  through (1,-1) 
hence y=-7x+6
they meet in(.78,.54)

.78-1=-.22
.54+1=1.54
hence$$\sqrt{\left( .22^{2}+1.54^{2} \right)}$$

anonymous · Answer

may b 1.55