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.

Question

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)

Answer 1

Are These Right??

I'm kinda suck on #30!!

Answer 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.