Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

Find the distance between the lines [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex].

[tex]\( t_1: y = -\frac{1}{2}x - \frac{1}{2} \)[/tex]

[tex]\( t_2: y = -\frac{1}{2}x + 1 \)[/tex]

Round your answer to the nearest tenth.

[tex]\(\square\)[/tex]


Sagot :

To determine the distance between two parallel lines in the form [tex]\( y = mx + b_1 \)[/tex] and [tex]\( y = mx + b_2 \)[/tex], we use the formula for the distance between two parallel lines:

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

First, let's write the equations of the lines in the general form [tex]\( ax + by + c = 0 \)[/tex]:

For line [tex]\( t_1 \)[/tex]:
[tex]\[ y = -\frac{1}{2}x - \frac{1}{2} \][/tex]
Rewriting this in the general form:
[tex]\[ -\frac{1}{2}x + y + \frac{1}{2} = 0 \quad \text{or} \quad -\frac{1}{2}x + y + \frac{1}{2} = 0 \][/tex]
So, [tex]\( a_1 = -\frac{1}{2} \)[/tex], [tex]\( b_1 = 1 \)[/tex], and [tex]\( c_1 = \frac{1}{2} \)[/tex].

For line [tex]\( t_2 \)[/tex]:
[tex]\[ y = -\frac{1}{2}x + 1 \][/tex]
Rewriting this in the general form:
[tex]\[ -\frac{1}{2}x + y - 1 = 0 \quad \text{or} \quad -\frac{1}{2}x + y - 1 = 0 \][/tex]
So, [tex]\( a_2 = -\frac{1}{2} \)[/tex], [tex]\( b_2 = 1 \)[/tex], and [tex]\( c_2 = -1 \)[/tex].

Now, applying the distance formula:
[tex]\[ \text{distance} = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \][/tex]

Substituting the values [tex]\( c_1 = \frac{1}{2} \)[/tex], [tex]\( c_2 = -1 \)[/tex], [tex]\( a = -\frac{1}{2} \)[/tex], and [tex]\( b = 1 \)[/tex]:
[tex]\[ \text{distance} = \frac{\left|\frac{1}{2} - (-1)\right|}{\sqrt{\left(-\frac{1}{2}\right)^2 + 1^2}} \][/tex]

Calculate the numerator:
[tex]\[ \left|\frac{1}{2} + 1\right| = \left|\frac{1}{2} + \frac{2}{2}\right| = \left|\frac{3}{2}\right| = \frac{3}{2} \][/tex]

Next, calculate the denominator:
[tex]\[ \sqrt{\left(-\frac{1}{2}\right)^2 + 1^2} = \sqrt{\frac{1}{4} + 1} = \sqrt{\frac{1}{4} + \frac{4}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \][/tex]

Finally, the distance is:
[tex]\[ \text{distance} = \frac{\frac{3}{2}}{\frac{\sqrt{5}}{2}} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5} \][/tex]

Simplifying this, we calculate:
[tex]\[ \frac{3\sqrt{5}}{5} \approx \sqrt{5} \times 0.6 \approx 2.1213203435596424 \][/tex]

Rounding to the nearest tenth:
[tex]\[ \text{distance} \approx 2.1 \][/tex]

Thus, the distance between the lines [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex] is approximately [tex]\( 2.1 \)[/tex].
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.