Respuesta :

Answer:

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

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:

[tex]d=\frac{|C-C'|}{\sqrt{A^{2} +B^{2} } }[/tex]

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:

[tex]d=\frac{|-22-(-15)|}{\sqrt{8^{2} +6^{2} } }[/tex]

Solving:

[tex]d=\frac{|-7|}{\sqrt{8^{2} +6^{2} } }[/tex]

[tex]d=\frac{7}{\sqrt{8^{2} +6^{2} } }[/tex]

[tex]d=\frac{7}{\sqrt{64+36 } }[/tex]

[tex]d=\frac{7}{\sqrt{100 } }[/tex]

[tex]d=\frac{7}{10}[/tex]

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

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