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avatar+118 

Find the distance between \((3,4)\) and the line \(4x+3y+7=0\).

 

Any help is greatly appreciated.

thx in advance!

 May 13, 2019
 #1
avatar+25824 
0

Here is one way:

4x+3y=-7

3y = -4x-7

y = -4/3 x- 7/3     so the slope is -4/3      Perpindicular slope is   3/4

 

Find line with 3/4 slope and point 3,4

4 = 3/4 (3) + b

b=  7/4

y = 3/4 x + 7/4

 

Now equate the two lines:

3/4x + 7/4 =   -4/3 x - 7/3

9/12 x + 7/4 = - 16/12 x - 7/3

25/12x = - 28/12 - 21/12

25/12 x = -49/12

x = -49/25

y= 196/75  - 7/3 = 196-7(25) /75 =  7/25

 

Now use distance formual between   3,4   and   -49/12 , 7/25

d^2 =   (3 -  - 49/25)^2     +    (4 - 7/25)^2

      =     124/25  ^2         +   93/25  ^2                  =    24025/625        take the square root   = 155/25 = 6.2

 

Here is another way:

     https://brilliant.org/wiki/dot-product-distance-between-point-and-a-line/

 May 13, 2019
 #2
avatar+111430 
+2

Here's a quick way......the distance between a point (a,b)  and a line in the form of

 

Ax + By + C  = 0  =

 

abs (A(a) + B(b) + C)

_________________       so we have        

sqrt (A^2 + B^2 )

 

abs ( 4(3) + 3(4) + 7 )                    31                    31     

___________________   =          ______ =         ___   = 6.2 units

sqrt (4^2 + 3^2 )                           sqrt(25)              5

 

 

cool cool cool

 May 14, 2019

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