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Find distance between parallel lines 3x+4 and 3x-6

Geometry question help please.

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Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
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The distance between two parallel lines y1=mx+b1 and y2=mx+b2 is given by
d= |b2-b1|/sqrt(m2+1).
With m=3, b1=4, and b2=-6, you get
d= |-6-4|/sqrt(32+1) = 10/√10 = √10.

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Thanks for the formula! I will remember it from now on. :-)
Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
4.9 4.9 (174 lesson ratings) (174)
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The line perpendicular to both of this lines has a slope of -1/3 and the length of a segment of it, which is between two those lines, gives the distance between them. Let us consider the line y=-x/3. It is perpendicular to both lines. It intercepts y=3x+4 at the point, which x-coordinate satisfies the following equation:
 
-x/3=3x+4; or x=-6/5=-1.2; y-coordinate is y=-1.2/-3=0.4
 
Analogously, it intercepts the line 3x-6 at x=1.8
 
-x/3=3x-6 x=9/5=1.8
 
y-coordinate is 1.8/-3=-0.6
 
Distance between two lines, 3x+4 and 3x-6, is:
 
d=√[(1.8-(-1.2))2+(-0.6-0.4)2]=√(32+(-1)2)=√10
 
Answer: √10