Question

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

Answer 1

Answer:

3x-4y= -7

3x4y= -5

may this helps you

Answer 2

Answer:

$d = \frac{c_1 -c_2}{ \sqrt{a²+b²} }$

$d=\frac{7-5}{ \sqrt{ {3}^{2} + { (- 4)}^{2} } }$

$d = \frac{2}{ \sqrt{9 + 16} }$

$d = \frac{2}{ \sqrt{25} }$

$d = \frac{2}{ \sqrt{5 \times 5} }$

$\boxed{d = \frac{2}{5} }$