Answer :
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].
[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].
