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# lines in coordinates

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

Sep 7, 2020

#3
+132
+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 8, 2020
edited by asinus  Sep 8, 2020
edited by asinus  Sep 8, 2020

#1
+10356
+1

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

Hello Guest!

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

$$d=4.4721$$

!

Sep 7, 2020
#2
+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
+132
+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

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