Question

The distance between the lines: 4x+3y-11 and 8x+6y=15 is​

Answer 1

Answer:

The distance between the lines: 4x+3y-11 and 8x+6y=15 is​ $\frac{7}{10}$

Step-by-step explanation:

You have two parallel and different lines. The form of the implicit equations is:

A*x + B*y + C= 0

A'*x + B'*y + C'=0

Since the two lines are parallel, the coefficients of their general equations must satisfy that the slopes are equal.

Therefore, the equations can be transformed so that the coefficients A and B are equal, multiplying or dividing one of them by a constant.

So, the distance between two lines can be expressed as follows:

$d=\frac{|C-C'|}{\sqrt{A^{2} +B^{2} } }$

In this case, you have:

• 4x+3y-11= 0 Multiplying this equation by 2 you get: 8x +6y -22=0
• 8x+6y-15 =0

Then, you have:

• A= 8
• B=6
• C= -22
• C'= -15

Replacing in the definition of distance:

$d=\frac{|-22-(-15)|}{\sqrt{8^{2} +6^{2} } }$

Solving:

$d=\frac{|-7|}{\sqrt{8^{2} +6^{2} } }$

$d=\frac{7}{\sqrt{8^{2} +6^{2} } }$

$d=\frac{7}{\sqrt{64+36 } }$

$d=\frac{7}{\sqrt{100 } }$

$d=\frac{7}{10}$

The distance between the lines: 4x+3y-11 and 8x+6y=15 is​ $\frac{7}{10}$