Asked by Bob
Given the line 4x - 3y = 12, what is the distance between it and the origin
I know the answer is 2.4 but how do u do it
I know the answer is 2.4 but how do u do it
Answers
Answered by
Steve
first way: The distance between (h,k) and Ax+By+C=0 is
|Ah+Bk+C|/√(A^2+B^2)
So, we have the point (0,0), giving
d = |12|/√(3^2+4^2) = 12/5 = 2.4
The other way is to realize that the shortest distance from a point to a line is along the line through the point and perpendicular to the given line.
Since our line has slope 4/3, the perpendicular line has slope -3/4.
SO, the normal line is y = -3/4 x
The two lines intersect where
-3/4 x = (4x-12)/3
That's at (48/25,-36/25)
The distance from there to (0,0) is
√((48/25)^2 + (36/25)^2) = 12/5 = 2.4
|Ah+Bk+C|/√(A^2+B^2)
So, we have the point (0,0), giving
d = |12|/√(3^2+4^2) = 12/5 = 2.4
The other way is to realize that the shortest distance from a point to a line is along the line through the point and perpendicular to the given line.
Since our line has slope 4/3, the perpendicular line has slope -3/4.
SO, the normal line is y = -3/4 x
The two lines intersect where
-3/4 x = (4x-12)/3
That's at (48/25,-36/25)
The distance from there to (0,0) is
√((48/25)^2 + (36/25)^2) = 12/5 = 2.4
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