Question

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

Answer 1

Explanation:

To find the distance between two lines, we can use the formula:

Distance = |(c2 - c1) / √(a^2 + b^2)|

where (a1, b1, c1) and (a2, b2, c2) are the coefficients of the two lines.

Given the equations of the lines as:

Line 1: 4x - 3y + 5 = 0

Line 2: 8 - 6y + 7 = 0

We can rewrite the equations in the form ax + by + c = 0:

Line 1: 4x - 3y - 5 = 0

Line 2: -6y + 15 = 0

Now, let's find the coefficients (a1, b1, c1) and (a2, b2, c2) for each line:

Line 1: a1 = 4, b1 = -3, c1 = -5

Line 2: a2 = 0, b2 = -6, c2 = 15

Plugging these values into the distance formula, we get:

Distance = |(c2 - c1) / √(a^2 + b^2)|

= |(15 - (-5)) / √((4^2 + (-3)^2))|

= |(15 + 5) / √(16 + 9)|

= |20 / √25|

= |20 / 5|

= 4

Therefore, the distance between the two lines 4x - 3y + 5 = 0 and 8 - 6y + 7 = 0 is 4 units.

Answer 2

The distance between the lines: 4x+3y-11 and 8x+6y=15 is​
`(7)/(10)`

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{\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{\sqrt{8^(2) +6^(2) } }`

`d=\frac{7}{\sqrt{8^(2) +6^(2) } }`

`d=(7)/(√(64+36 ) )`

`d=(7)/(√(100 ) )`

`d=(7)/(10)`

The distance between the lines: 4x+3y-11 and 8x+6y=15 is​
`(7)/(10)`