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Please help with this: Find the distance between  (3,4) and the line 4x+3y+7=2x-y+5.

 Mar 9, 2022
 #1
avatar+122390 
+1

Rearrange  the  equation   in  the form    Ax + By + C  =  0

 

2x + 4y + 2    =  0

 

We have  the following  for the distance from  (x, y)  to  the line   =

 

l Ax  + By + C l                         l 2 (3)  + 4 ( 4)  + 2  l               24               24

________________  =         ___________________   =    ______  =   ______  =

  sqrt [ A^2 + B^2]                     sqrt  [ 2^2 + 4^2 ]                 sqrt 20        2 sqt 5

 

 

12 / sqrt 5   =   

 

(12/5) sqrt (5)

 

See the graph here :  https://www.desmos.com/calculator/3dgerbkzta

 

cool cool cool

 Mar 9, 2022
 #2
avatar+36417 
+1

Another (but longer) method ( because I can never remember the equation for the minumum distance between a line and a point)

 

4x+3y+7=2x-y+5     re-arrange to y = -1/2 x -1/2

Now use distance formula

d^2   =  (x -3)^2  + (4 - (-1/2x-1/2))^2

             x^2 -6x + 9    +  1/4x^2 + 4.5 x + 20.25

  d^2        =  5/4 x^2 - 1.5 x + 29.25          <====== take derivative and set = 0

                  10/4 x -1.5 = 0

                       x = 0.6            <====== sub this value into the line equation to find y = - 0.8

 

NOW, use the distance formula between (0.6, -0.8)   and (3,4)

        d^2 = ( 3 - 0.6)2 + (4- - 0.8)2 

         d^2 = 28.8

           d = sqrt (28.8)             ( which = same answer that chris found....)   

 Mar 9, 2022

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