Question

find distance between parallel lines 2x-3y+7=0 & 2x-3y-6=0.​

Answer 1

Answer:

on compairing these equations with Ax+By+C1=0 and Ax+By+C2=0,we get A=2 B=-3 C1=7C2=-6

distance between 2 parallel lines is:

d=|C1-C2|/√Asquare +Bsquare

d=|7-(-6)|/√2×2+(-3)(-3),d=|13|/√4+9

d=13/√13,on rationalising d=13×√13/√13×√13

d=13√13/13,d=√13units

Answer 2

Answer:

Distance between parallel lines 2x-3y+7=0 & 2x-3y-6=0 is $\sqrt{13}$ units

Step-by-step explanation:

Given data   2x-3y+7=0  

                      2x-3y-6=0  are two parallel lines

here we need to find distance between the given parallel lines

⇒ compare the given line with ax+by+c₁ = 0 and ax+by+c₂ = 0

⇒ from given parallel lines a = 2, b = -3 and c₁ = 7  c₂ -6

⇒ the formula for distance between two parallel lines is given by

                                       d = |c₁-c₂|/√(a²+ b²)

                                          = l 7 - (-6) l /√(2²+(-3)²)

                                          = l 7 +6 l /√(4 + 9 )  

                                          = 13/ $\sqrt{13}$  = $\frac{\sqrt{13}\sqrt{13} }{\sqrt{13} }$ = $\sqrt{13}$ units

⇒ distance between parallel lines 2x-3y+7=0 and 2x-3y-6=0  is $\sqrt{13}$ units