Question

28. The perpendicular distance between two
parallel lines 3x + 4y - 6 = 0 and
6x + 8y + 7 = 0 is equal to
(A) 19/5 unit (B) 19/10 unit
(C) 19/2 unit
(D) 10/19 unit​

Answer 1

Answer:

option (b) is correct

Step-by-step explanation:

Formula used:

The distance between the parallel lines

$ax+by+c_1=0\:and\:ax+by+c_2=0\:is\:|\frac{c_1-c_2}{\sqrt{a^2+b^2}}|$

Given lines are

$3x+4y-6=0$

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

Here, a=3, b=4, $c_1=-6,\:c_2=\frac{7}{2}$

The perpendicular distance between the given two parallel lines

$=|\frac{c_1-c_2}{\sqrt{a^2+b^2}}|$

$=|\frac{-6-\frac{7}{2}}{\sqrt{3^2+4^2}}|$

$=|\frac{\frac{-12-7}{2}}{\sqrt{9+16}}|$

$=|\frac{\frac{-19}{2}}{\sqrt{25}}|$

$=|\frac{\frac{-19}{2}}{5}|$

$=|\frac{-19}{10}|$

$=\frac{19}{10}$ units

Answer 2

Answer:

The correct answer is B) 19/10.

Step-by-step explanation:

Let us consider the equation 3x + 4y - 6 = 0 be A1.

Let us consider the equation 6x + 8y + 7 = 0 be A2.

We know that,

Slope = - (coefficient of y / coefficient of x).

So,

Slope of A1 = - 4/3

Slope of A2 = - 8/6 = - 4/3

Here we can see that the slope of A1 is equal to the slope of A2. This means that both the lines are parallel as mentioned in the question.

The equation 3x + 4y - 6 = 0 can also be written as 3x + 4y = 6.

Here c1 = 6.

The equation 6x + 8y + 7 = 0 can also be written as 3x +4y = -7/2.

here c2 = - 7/2 .

Distance between parallel lines is =( | c1 - c2 |) / (√(a² + b²)) .

                                                         = ( |6 + 7/2| ) / ( √(3² + 4²))

                                                          = 9.5 / 5

                                                           = 19 / 10.