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)
b.(-5, 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.

Answers

Answer 1

Answer: 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

Answer 2

Answer:

Answer: The answers are given below.

Step-by-step explanation: The calculations are as follows:

(1). The mid-point of a line segment divides it in the ratio m : n = 1 : 1.

So, the co-ordinates of the mid-point of the segment joining the points (a, b) and (j, k) are

$\left((mj+na)/(m+n),(mk+nb)/(m+n)\right)\n\n\n=\left((* j+1* a)/(1+1),(1* k+1* b)/(1+1)\right)\n\n\n=\left((j+a)/(2),(k+b)/(2)\right).$

So, the co-ordinates of the mid-point are $\left((j+a)/(2),(k+b)/(2)\right).$

Thus, (d) is the correct option.

(2). The co-ordinates of the points 'T' and 'J' are (0, 4) and (0, 2) respectively.

Let, (a, b) be the co-ordinates of the point 'H'.

Since 'T' is the mid-point of the line segment JH, so we have

$(0,4)=\left((0+a)/(2),(2+b)/(2)\right)\n\n\n\Rightarrow (0,4)=\left((a)/(2),(2+b)/(2)\right)\n\n\n\Rightarrow (a)/(2)=0,~~~(2+b)/(2)=4\n\n\n\Rightarrow a=0,~~\Rightarrow 2+b=8~~~\Rightarrow b=6.$