_{1}=mx+b

_{1}and y

_{2}=mx+b

_{2}is given by

_{2}-b

_{1}|/sqrt(m

^{2}+1).

_{1}=4, and b

_{2}=-6, you get

^{2}+1) = 10/√10 = √10.

Geometry question help please.

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The distance between two parallel lines y_{1}=mx+b_{1} and y_{2}=mx+b_{2} is given by

d= |b_{2}-b_{1}|/sqrt(m^{2}+1).

With m=3, b_{1}=4, and b_{2}=-6, you get

d= |-6-4|/sqrt(3^{2}+1) = 10/√10 = √10.

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}]=√(3^{2}+(-1)^{2})=√10

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