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Find the distance between the lines y = 1/2*x + 1 and y = 1/2*x - 4.

 Sep 7, 2020

Best Answer 

 #3
avatar+132 
+1

I made this answer, before creating my account. How do I claim this answer?

Much easier:

 

Note that \(d=\frac{c_2-c_1}{\sqrt{A^2+B^2}}\) for lines \(ax+by+c_1=0, ax+by+c_2=0\). So we have a=-1/2, b=1, c1=-1, c2=4. Plugging, we get \(\frac{4+1}{\sqrt{1/4+1}}=\frac{5}{\sqrt{5/4}}=\frac{5}{\frac{\sqrt{5}}{2}}=\frac{10}{\sqrt{5}}=\frac{10\sqrt{5}}{5}=\boxed{2\sqrt{5}}\) . 

 

Note: This yields the same answer as @asinus got, just in exact form

 Sep 8, 2020
edited by asinus  Sep 8, 2020
edited by asinus  Sep 8, 2020
 #1
avatar+10356 
+1

Find the distance d between the lines y = 1/2*x + 1 and y = 1/2*x - 4.

 

Hello Guest!

 

\(tan\ \alpha=\frac{1}{2}\\ \alpha=atan\ \frac{1}{2}\\ d=(1-(-4))\cdot cos\ \alpha\\ d=(1-(-4))\cdot cos\ atan\ \frac{1}{2}\)

\(d=4.4721\)

laugh  !

 Sep 7, 2020
 #2
avatar
+1

Much easier:

 

Note that \(d=\frac{c_2-c_1}{\sqrt{A^2+B^2}}\) for lines \(ax+by+c_1=0, ax+by+c_2=0\). So we have a=-1/2, b=1, c1=-1, c2=4. Plugging, we get \(\frac{4+1}{\sqrt{1/4+1}}=\frac{5}{\sqrt{5/4}}=\frac{5}{\frac{\sqrt{5}}{2}}=\frac{10}{\sqrt{5}}=\frac{10\sqrt{5}}{5}=\boxed{2\sqrt{5}}\) . 

 

Note: This yields the same answer as @asinus got, just in exact form

 Sep 7, 2020
edited by asinus  Sep 8, 2020
 #3
avatar+132 
+1
Best Answer

I made this answer, before creating my account. How do I claim this answer?

Much easier:

 

Note that \(d=\frac{c_2-c_1}{\sqrt{A^2+B^2}}\) for lines \(ax+by+c_1=0, ax+by+c_2=0\). So we have a=-1/2, b=1, c1=-1, c2=4. Plugging, we get \(\frac{4+1}{\sqrt{1/4+1}}=\frac{5}{\sqrt{5/4}}=\frac{5}{\frac{\sqrt{5}}{2}}=\frac{10}{\sqrt{5}}=\frac{10\sqrt{5}}{5}=\boxed{2\sqrt{5}}\) . 

 

Note: This yields the same answer as @asinus got, just in exact form

edited by asinus  Sep 8, 2020
edited by asinus  Sep 8, 2020

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