Question

The penpendicular distance between two Parallel lines
3x+4y-6= 0 and 6x+8y+ 7 =0 is equal to​

Answer 1

Answer:

Formula used:

The distance between the parallel lines

ax+by+c_1=0\:and\:ax+by+c_2=0\:is\:|\frac{c_1-c_2}{\sqrt{a^2+b^2}}|ax+by+c1=0andax+by+c2=0is∣a2+b2c1−c2∣

Given lines are

3x+4y-6=03x+4y−6=0

3x+4y+\frac{7}{2}=03x+4y+27=0

Here, a=3, b=4, c_1=-6,\:c_2=\frac{7}{2}c1=−6,c2=27

The perpendicular distance between the given two parallel lines

=|\frac{c_1-c_2}{\sqrt{a^2+b^2}}|=∣a2+b2c1−c2∣

=|\frac{-6-\frac{7}{2}}{\sqrt{3^2+4^2}}|=∣32+42−6−27∣

=|\frac{\frac{-12-7}{2}}{\sqrt{9+16}}|=∣9+162−12−7∣

=|\frac{\frac{-19}{2}}{\sqrt{25}}|=∣252−19∣

=|\frac{\frac{-19}{2}}{5}|=∣52−19∣

=|\frac{-19}{10}|=∣10−19∣

=\frac{19}{10}=1019 units