Precalc

How do you know what quadrant these two are in? I'm having a little trouble determining. Can someone check if my answers are correct for all and help me fix it to make it right.

10.) The point is on the x-axis and 12 units to the left of the y-axis.
My answer: (0, -12)

12.) y<-5 = Quadrant 3

16.) xy<0 = Quadrant 1

26.) (-3, -1) & (2,-1)
Find distance.
My answer: 5

30.) the 3 points of the triangle they gave are (1,5), (5,-2), & (1,-2)
Find the length of each side of the right triangle & show that these lengths satisfy the pythagorean theorem.

??

42.) (-1,2) & (5,4)
Distance answer= square root 40
midpoint= 2,3)

47.) (-16.8, 12.3) & (5.6, 4.9)
Distance answer= square root 556.52 or 23.6
Midpoint= (-5.6, 8.6)

asked by Josh
  1. Are These Right??

    I'm kinda suck on #30!!

    posted by Josh
  2. 10. Any point on the x-axis has a Y of 0;
    12 units to the left of the Y-axis is an
    X value (-)12: P(-12,0).

    12. Q3, or Q4 depending on whether X is
    positive or negative.

    16. X,Y < 0. Q3.

    26. (-3,-1), (2,-1).
    D^2 = X^2 + Y^2 = (2+3)^2 + (-1+1)^2 =
    25 + 0 = 25.
    D = 5.

    30. (1,5), (5,-2), (1,-2).
    First, sketch the triangle showing the
    given points.

    The vertical side: (1,5), (1,-2).
    L^2 = X^2 + Y^2 = (1-1)^2 + (-2-5)^2 =
    0 + 49 = 49.
    L = 7. or 5-(-2) = 5 + 2 = 7.

    Horizontal side: (1,-2), (5,-2).
    L = 5-1 = 4.

    Hyp : (1,5), (5,-2).
    L^2 = (5-1)^2 + (-2-5)^2 =

    Use the length of the hor. and ver. sides to calculate length of hyp.

    L^2 = X^2 + Y^2 .

    42. (-1,2),(x,y), (5,4).
    D^2 = (5+1)^2 + (4-2)^2

    Mid-point:
    (5+1) = 2(x+1).
    Solve for X.

    (4-2) = 2(y-2).
    Solve for Y.

    47. Same procedure as #42.





    posted by Henry

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