Question

The distance between the parallel lines 6x + 8y + 21 = 0 and 3x + 4y + 7 = 0 is​

Answer 1

Answer:

Hope my answer will help you.

Answer 2

Answer:

Distance between the given parallel lines = 0.7  units

Step-by-step explanation:

given, lines 6x + 8y + 21 = 0 and 3x + 4y + 7 = 0 are parallel.

=> their slopes will be equal

consider, 6x+8y+21=0

              => 8y = -6x-21

              => y = $\frac{-6}{8}$x + $\frac{-21}{8}$

              => y = $\frac{-3}{4}$x + $\frac{-21}{8}$  --------- (1)

similarly,  3x + 4y + 7 = 0

              => 4y = -3x-7

              => y = $\frac{-3}{4}$x + $\frac{-7}{4}$  --------- (2)

from (1) and (2) we have the slopes, (co-effecient of x) are equal

ie., m₁ = m₂ = $\frac{-3}{4}$

=> the 2 lines are parallel

also, c₁ = $\frac{-21}{8}$ and c₂ = $\frac{-7}{4}$

we know for 2 parallel lines, y=mx+c₁ & y = mx+c₂ , the distance between them is given by the formula,

  d = $\frac{| (c_1-c_2)|}{\sqrt{1+m^2} }$ units

so, here,

d = $\frac{| (\frac{-21}{8}) - (\frac{-7}{4})|}{\sqrt{1+(\frac{-3}{4} )^2} }$ units

   = $\frac{| (\frac{-21}{8}) + (\frac{14}{8})|}{\sqrt{1+(\frac{9}{16} )} }$ units

   = $\frac{| \frac{(-21+14)}{8}|}{\sqrt{\frac{16+9}{16} } }$ units

   = $\frac{| \frac{(-7)}{8}|}{\sqrt{\frac{25}{16} } }$ units

   = $\frac{ \frac{(7)}{8}}{\frac{5}{4} }$ units

   = $\frac{7}{8}$ * $\frac{4}{5}$ units

   = $\frac{7}{2}$ * $\frac{1}{5}$ units

   = $\frac{7}{10}$  units

   = 0.7  units