Precalc
posted by Josh
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 xaxis and 12 units to the left of the yaxis.
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)

Josh
Are These Right??
I'm kinda suck on #30!! 
Henry
10. Any point on the xaxis has a Y of 0;
12 units to the left of the Yaxis 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 = (11)^2 + (25)^2 =
0 + 49 = 49.
L = 7. or 5(2) = 5 + 2 = 7.
Horizontal side: (1,2), (5,2).
L = 51 = 4.
Hyp : (1,5), (5,2).
L^2 = (51)^2 + (25)^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 + (42)^2
Midpoint:
(5+1) = 2(x+1).
Solve for X.
(42) = 2(y2).
Solve for Y.
47. Same procedure as #42.
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