CK-12 Geometry-Concepts

(Marvins-Underground-K-12) #1

3.10. Distance Formula in the Coordinate Plane http://www.ck12.org



  1. Find the slope of the perpendicular line. The opposite reciprocal of−^12 is 2. We know the perpendicular line has
    a slope of 2 and contains the point (2, -5), so− 5 = 2 ( 2 )+band thereforeb=−9. Next, we need to figure out where
    the linesy=−^12 x+1 andy= 2 x−9 intersect:


2 x− 9 =−

1


2


x+ 1
2. 5 x= 10
x= 4 ,and thereforey=− 1

So the lines intersect at the point (4, -1). Now, use the distance formula to find the distance between (4, -1) and (2,


-5):



( 2 − 4 )^2 +(− 5 −(− 1 ))^2 =



4 + 36 =



40 ≈ 6. 32 units.

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Find the distance between each pair of points. Round your answer to the nearest hundredth.



  1. (4, 15) and (-2, -1)

  2. (-6, 1) and (9, -11)

  3. (0, 12) and (-3, 8)

  4. (-8, 19) and (3, 5)

  5. (3, -25) and (-10, -7)

  6. (-1, 2) and (8, -9)

  7. (5, -2) and (1, 3)

  8. (-30, 6) and (-23, 0)


Determine the shortest distance between the given line and point. Round your answers to the nearest hundredth.


9.y=^13 x+4;( 5 ,− 1 )
10.y= 2 x−4;(− 7 ,− 3 )
11.y=− 4 x+1;( 4 , 2 )
12.y=−^23 x−8;( 7 , 9 )


  1. The distance between two points is 5 units. One point is (2, 2). How many possibilities are there for the second
    point? You may assume that the second point is made up of integers.

  2. What if in the previous question the second point was not necessarily made up of integers? What shape would
    be created by plotting all of the possibilities for the second point?

  3. Find one possibility for the equation of a line that is exactly 6 units away from the point (-3,2).

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