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  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)

  1. 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.

Verified Answer :

nobillionaireNobley
  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:  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

\(\left(\dfrac{mj+na}{m+n},\dfrac{mk+nb}{m+n}\right)\\\\\\=\left(\dfrac{\times j+1\times a}{1+1},\dfrac{1\times k+1\times b}{1+1}\right)\\\\\\=\left(\dfrac{j+a}{2},\dfrac{k+b}{2}\right).\)

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

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

\((0,4)=\left(\dfrac{0+a}{2},\dfrac{2+b}{2}\right)\\\\\\\Rightarrow (0,4)=\left(\dfrac{a}{2},\dfrac{2+b}{2}\right)\\\\\\\Rightarrow \dfrac{a}{2}=0,~~~\dfrac{2+b}{2}=4\\\\\\\Rightarrow a=0,~~\Rightarrow 2+b=8~~~\Rightarrow b=6.\)

So, the co-ordinates of 'H' are (0, 6).

Thus, (a) is the correct option.

(3). The given point is (-4, 3).

Since 'x' co-ordinate is negative and 'y' co-ordinate is positive, so the given point lies in Quadrant II.

Thus, (b) is the correct option.

(4). The given point is (6, 0).

Here, 'y' co-ordinate is zero, so the point lies on the X-axis.

Thus, (a) is the correct option.

(5). The slope-intercept form of a line is

y = mx + c, where, 'm' is the slope and 'c' is the y-intercept.

If slope, m = 0, then the equation becomes

y = 0 × m + c

⇒ y = c, which is the equation of a line parallel to X-axis.

Thus, option (a) is correct.

(7). The point (a, 3) lies on the graph of the equation 5x + y = 8, so we have

\(5x+y=8\\\\\Rightarrow 5\times a+3=8\\\\\Rightarrow 5a=5\\\\\Rightarrow a=1.\)

Thus, (a) is the correct option.

(8). Given that the equation of a circle is

\((x+5)^2+(y-7)^2=36~~~~~~~~~~~~~~~~~~~~~~~~~(i)\)

We know that the equation of a circle with centre (g, h) and radius 'r' units is given by

\((x-g)^2+(y-h)^2=r^2.\)

Comparing equation (i) with the above general equation, we get

(g, h) = (-5, 7).

So, the centre point is (-5, 7).

Thus, option (b) is correct.

(9).  Given that the equation of a circle is

\((x+5)^2+(y-7)^2=36~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\)

We know that the equation of a circle with centre (g, h) and radius 'r' units is given by

\((x-g)^2+(y-h)^2=r^2.\)

Comparing equation (ii) with the general equation, we get

r² = 36   ⇒ r = 6 units.

So, the radius is 6 units.

Thus, option (a) is correct.

(10). Given that the equation of a circle is

\((x-2)^2+(y-6)^2=4~~~~~~~~~~~~~~~~~~~~~~~~~(iii)\)

We know that the equation of a circle with centre (g, h) and radius 'r' units is given by

\((x-g)^2+(y-h)^2=r^2.\)

Comparing equation (iii) with the above general equation, we get

(g, h) = (2, 6).

So, the centre point is (-5, 7). Hence, the given statement is TRUE.

All the questions are ANSWERED.

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