Question

Find the distance between the parallel lines 4x-3y+5=0 and 4x-3y+7=0​

Answer 1

Answer:

distance =|7-5|/√(16+9)=2/5

Answer 2

Answer:

$we \: have \: \\ \\ 4x - 3y + 5 = 0 \\ \\ 4x - 3y + 7 = 0 \\ \\ \\ by \: distance \: formula \: \\ \\ d = \frac{c2 - c1}{ \sqrt{ {a}^{2} + {b}^{2} } } \\ \\ d = \frac{7 - 5}{ \sqrt{16 + 9} } \\ \\ d = \frac{2}{5}$