1. the midpoint of the segment joining points (a, b) and ( j, k) is: a. (j-a, k-b) b. ((j-a)/2,(k-b)/2) c. (j+a, k+b) d. ((j+a)/2,(k+b)/2)
2. point t is the midpoint of jh . the coordinate of t is (0, 4) and the coordinate of j is (0, 2). the coordinate of h is: a.(0, 6)
b.(0, 3)
c.(0, 7)
d.(0, 11)
3. what about this one point (-4, 3) lies in quadrant
a. i
b. ii
c. iii
d. iv 4. point (6, 0) lies on the.
a. x-axis
b. y-axis
c. line y = x 5. any line with no slope is parallel to the
a. x-axis
b. y-axis
c. line y = x 7. if the point (a,3) lies on the graph of the equation 5x + y = 8, then a=
a.1
b.-1
c.-7 8. if the equation of a circle is (x + 5)^2 + (y - 7)^2 = 36, its center point is
a.(5, 7)
, 7)
c.(5, -7) 9. if the equation of a circle is (x + 5)^2 + (y - 7)^2 = 36, its radius is
a.6
b.16
c.36 10. if the equation of a circle is (x - 2)^2 + (y - 6)^2 = 4, the center is point (2, 6).
true or false.

1. The midpoint of the segment joining points (a, b) and ( j, k) is ((j+a)/2,(k+b)/2)

2. Let the coordinate of H be (a, b)
T(0, 4) = ((a + 0)/2, (b + 2)/2)
(a + 0)/2 = 0 => a + 0 = 0 => a = 0
(b + 2)/2 = 4 => b + 2 = (2 x 4) = 8 => b = 8 - 2 = 6
Therefore, the cordinate of H is (0, 6)

3. Point (-4, 3) lies in Quadrant II

4. Point (6, 0) lies on the x-axis

5. Any line with no slope is parallel to the y-axis

7. a is the value of the x-coordinate.
5a + 3 = 8
5a = 8 - 3 = 5
a = 5/5 = 1
a = 1

8. Equation of a circle is given by (x - a)^2 + (y - b)^2 = r^2; where (a, b) is the center and r is the radius.
For the given circle (x + 5)^2 + (y - 7)^2 = 36 => (x - (-5))^2 + (y - 7)^2 = 6^2 => a = -5 and b = 7.
Therefore, its center point is (-5, 7)

9. Equation of a circle is given by (x - a)^2 + (y - b)^2 = r^2; where (a, b) is the center and r is the radius.
For the given circle (x + 5)^2 + (y - 7)^2 = 36 => (x - (-5))^2 + (y - 7)^2 = 6^2 => r = 6.
Therefore, its radius is 6

10. If the equation of a circle is (x - 2)^2 + (y - 6)^2 = 4, the center is point (2, 6).
True

The answer is -pie over 2