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Hero
 2 years ago
Best ResponseYou've already chosen the best response.0Hint: Use the distance formula for lines

ChmE
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large d=\sqrt{(x_2x_1)^2+(y_2y_1)^2}\]

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0There's two ways to do it. You can either use the distance formula for LINES (yes there is such a thing) or you can find the line perpendicular to both given lines since they are parallel.

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0It's just a formula bro. It's nothing to get overly eager and excited about.

kenneyfamily
 2 years ago
Best ResponseYou've already chosen the best response.0ok the question is the distance is 4/5. true or false?

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0The formula is \[ d = \frac{ax + by + c}{\sqrt{a^2 + b^2}}\] Where (x,y) is a point on one of the lines and a, b, c are the coefficients of the other line.

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0And I should say that this is specifically the formula for the distance between two PARALLEL lines.

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0(x,y) = (.5,0) is a point on the first given line. So use a = 6, b = 8, c = 5 for the other one.

kenneyfamily
 2 years ago
Best ResponseYou've already chosen the best response.0how would you non parallel lines? @Hero such as x + y  1 = 0 and y = x + 6

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0If the lines are not parallel, then the distance between them would not be consistent.

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0That formula I gave you (as I stated previously) is for PARALLEL lines only

kenneyfamily
 2 years ago
Best ResponseYou've already chosen the best response.0hmm but it is answerable

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0Well, actually, what you just posted are two parallel lines.
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