Question

Find the perpendicular distance between the lines 3x+4y-5=0 and 6x+8y-45=0​

Answer 1

The perpendicular distance between the lines $3x+4y-5=0\ and\ 6x+8y-45=0$ is 3.5.

Step-by-step explanation:

Given:

The equations of the two lines are given as:

$3x+4y-5=0\\6x+8y-45=0$

Let us first find the slopes of the two equations.

Rewriting the above equations in the form

$y=mx+c\\Where,m\rightarrow slope\ of\ line\\c\rightarrow y-intercept$

$3x+4y-5=0\\3x+4y=5\\4y=-3x+5\\y=-\frac{3}{4}x+\frac{5}{4}$

Slope of the line is $-\dfrac{3}{4}$

$6x+8y-45=0\\6x+8y=45\\8y=-6x+45\\y=-\frac{6}{8}x+\frac{45}8}\\y=-\frac{3}{4}x+{45}{8}$

Slope of the line is $-\dfrac{3}{4}$

Therefore, the slopes of both the lines are same. So, the lines are parallel to each other.

Now, for parallel lines $Ax+By+C_1=0\ and\ Ax+By+C_2=0$, the perpendicular distance is equal to:

$d=\frac{|C_2-C_1|}{\sqrt{A^2+B^2}}$

Here, $A=3,B=4,C_1=-5,C_2=-\frac{45}{2}$

Now, applying the formula, we get the distance as:

$d=\frac{|-\frac{45}{2}-(-5)|}{\sqrt{3^2+4^2}}\\\\d=\frac{|-\frac{45}{2}+5|}{\sqrt{9+16}}\\\\d=\frac{|\frac{-45+10}{2}|}{\sqrt{25}}\\\\d=\frac{\frac{35}{2}}{5}\\\\d=\frac{35}{10}=3.5$

Therefore, the perpendicular distance between the lines $3x+4y-5=0\ and\ 6x+8y-45=0$ is 3.5.

Answer 2

Given : parallel lines 3x + 4y = 5 and   6x + 8y = 45  

To Find  :  Distance between parallel lines

Solution :

3x + 4y = 5  

6x + 8y = 45

Slope of the line =  - 3/4

Hence Slope of perpendicular line =  4/3

Let say a line  y = 4x/3    cuts both the line

3x + 4y = 5  

=> 3x + 4(4x/3) = 5

=> 9x + 16x  = 15

=> 25x  = 15

=> x  = 3/5

    y = 4/5

6x + 8y = 45

=> 6x +8(4x/3) = 45

=> 18x + 32x  = 135

=> 50x  = 135

=> 10x = 27

=> x  = 27/10

    y =  36/10

Points are ( 3/5 , 4/5)  and ( 27/10 , 36/10)

Distance = √(27/10 - 3/5)²  + (36/10 - 4/5)²

= (1/10)√ 21² + 28²

= (7/10)√3² + 4²

= (7/10)5

= 35/10

= 3.5

perpendicular distance between the lines 3x+4y-5=0 and 6x+8y-45=0​ = 3.5

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