Question

2) Distance between the lines 3y - 2x – 4 = 0 and 4x - 6y + 7 = 0 is
15
2713
1
213
15
13
1
V13​

Answer 1

Given : 3y - 2x – 4 = 0 and 4x - 6y + 7 = 0  

To Find :  Distance between lines 3y - 2x – 4 = 0 and 4x - 6y + 7 = 0  

Solution:

3y - 2x – 4 = 0 and 4x - 6y + 7 = 0  

=> Slope =  2/3

Slope of perpendicular line = - 3/2

Hence y = -3x/2  + c

=>  3x  + 2y  = k    

3y - 2x – 4 = 0  =>  3y - 2x = 4

x = - 2 ,  y = 0  is one point  (-2 , 0)  on 3y - 2x = 4

lets draw perpendicular line 3x  + 2y  = k   passing through this

=> 3(-2) + 2(0) = k

=> k = - 6

=> 3x  + 2y  = -6

 3x  + 2y  = -6

 4x - 6y + 7 = 0  

3 (  3x  + 2y  ) +   4x - 6y + 7 = 3(-6) + 0

=> 13x + 7 = -18

=> x = -25/13

   y = -3/26

 

(-2 , 0)   and ( -25/13 , - 3/26)

Distance =  √(-25/13  + 2)²+ (-3/26 - 0)²

= √13 / 26

= 1/2√13

Distance between lines 3y - 2x – 4 = 0 and 4x - 6y + 7 = 0   is 1/2√13

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