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Find the distance between the lines [tex]\(\ell_1\)[/tex] and [tex]\(\ell_2\)[/tex].

[tex]\[
\ell_1: -x + y = 0
\][/tex]
[tex]\[
\ell_2: -2x + 2y = 7
\][/tex]

Round your answer to the nearest tenth.


Sagot :

To find the distance between the two parallel lines, we start by rewriting their equations into a more standard form.

For line [tex]\( \ell_1 \)[/tex]:
[tex]\[ -x + y = 0 \][/tex]
This can be rewritten as:
[tex]\[ y = x \][/tex]

For line [tex]\( \ell_2 \)[/tex]:
[tex]\[ -2x + 2y = 7 \][/tex]
We can divide through by 2 to simplify:
[tex]\[ -x + y = 3.5 \][/tex]
So, this can be rewritten as:
[tex]\[ y = x + 3.5 \][/tex]

Now we have the two lines in slope-intercept form:
[tex]\[ \ell_1: y = x \][/tex]
[tex]\[ \ell_2: y = x + 3.5 \][/tex]

Since these lines are parallel, they have the same slope (m = 1). To find the distance between two parallel lines of the form [tex]\( y = mx + c_1 \)[/tex] and [tex]\( y = mx + c_2 \)[/tex], we can use the formula:

[tex]\[ \text{distance} = \frac{|c_2 - c_1|}{\sqrt{1 + m^2}} \][/tex]

For these lines:
[tex]\[ c_1 = 0 \quad \text{(from } y = x \text{)} \][/tex]
[tex]\[ c_2 = 3.5 \quad \text{(from } y = x + 3.5 \text{)} \][/tex]
[tex]\[ m = 1 \][/tex]

Substitute these values into the distance formula:
[tex]\[ \text{distance} = \frac{|3.5 - 0|}{\sqrt{1 + 1^2}} \][/tex]

Simplify the expression:
[tex]\[ \text{distance} = \frac{3.5}{\sqrt{2}} \][/tex]

We know that [tex]\( \sqrt{2} \approx 1.414 \)[/tex], so:
[tex]\[ \text{distance} \approx \frac{3.5}{1.414} \][/tex]
[tex]\[ \text{distance} \approx 2.4748737341529163 \][/tex]

Rounding to the nearest tenth:
[tex]\[ \text{distance} \approx 2.5 \][/tex]

So, the distance between line [tex]\( \ell_1 \)[/tex] and line [tex]\( \ell_2 \)[/tex] is approximately 2.5 units.
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