To find the possible values of [tex]\( b \)[/tex] such that the distance between the points [tex]\( P(12, 8) \)[/tex] and [tex]\( Q(6, b) \)[/tex] is 10 units, we will use the distance formula. Let's go through the steps to solve this problem.
### Step-by-Step Solution
#### Step 1: Write down the distance formula
The distance [tex]\( d \)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] in a Cartesian plane is given by:
[tex]\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\][/tex]
#### Step 2: Substitute the given coordinates and distance
Here, point [tex]\( P \)[/tex] has coordinates [tex]\((x_1, y_1) = (12, 8)\)[/tex], and point [tex]\( Q \)[/tex] has coordinates [tex]\((x_2, y_2) = (6, b)\)[/tex]. The distance [tex]\( d \)[/tex] is given as 10. So we substitute these values into the distance formula:
[tex]\[
10 = \sqrt{(6 - 12)^2 + (b - 8)^2}
\][/tex]
#### Step 3: Simplify inside the square root
Calculate [tex]\((6 - 12)^2\)[/tex]:
[tex]\[
6 - 12 = -6 \quad \text{and} \quad (-6)^2 = 36
\][/tex]
Substitute back into the distance formula:
[tex]\[
10 = \sqrt{36 + (b - 8)^2}
\][/tex]
#### Step 4: Eliminate the square root by squaring both sides
Square both sides of the equation to remove the square root:
[tex]\[
10^2 = (36 + (b - 8)^2)
\][/tex]
[tex]\[
100 = 36 + (b - 8)^2
\][/tex]
#### Step 5: Isolate the squared term
Subtract 36 from both sides:
[tex]\[
100 - 36 = (b - 8)^2
\][/tex]
[tex]\[
64 = (b - 8)^2
\][/tex]
#### Step 6: Solve for [tex]\( b \)[/tex] by taking the square root of both sides
Take the square root of both sides to solve for [tex]\( b - 8 \)[/tex]:
[tex]\[
\sqrt{64} = |b - 8|
\][/tex]
[tex]\[
8 = |b - 8|
\][/tex]
Since the absolute value [tex]\( |b - 8| = 8 \)[/tex], there are two solutions:
[tex]\[
b - 8 = 8 \quad \text{or} \quad b - 8 = -8
\][/tex]
#### Step 7: Solve the resulting equations
Solve for [tex]\( b \)[/tex] in both cases:
[tex]\[
b - 8 = 8 \quad \Rightarrow \quad b = 16
\][/tex]
[tex]\[
b - 8 = -8 \quad \Rightarrow \quad b = 0
\][/tex]
#### Step 8: Conclude the possible values of [tex]\( b \)[/tex]
The possible values of [tex]\( b \)[/tex] are:
[tex]\[
b = 16 \quad \text{or} \quad b = 0
\][/tex]
So the possible values of [tex]\( b \)[/tex] such that the distance between the points [tex]\( P(12, 8) \)[/tex] and [tex]\( Q(6, b) \)[/tex] is 10 units are [tex]\( b = 0 \)[/tex] and [tex]\( b = 16 \)[/tex].
