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Certainly! Let's solve the problem step by step.
Problem: Find [tex]\( x \)[/tex] such that the distance between points [tex]\( P(5, -3) \)[/tex] and [tex]\( Q(0, 1) \)[/tex] is equal to the distance between points [tex]\( Q(0, 1) \)[/tex] and [tex]\( R(x, 6) \)[/tex].
Step-by-Step Solution:
1. Calculate the distance [tex]\( PQ \)[/tex]:
The distance [tex]\( PQ \)[/tex] between points [tex]\( P \)[/tex] and [tex]\( Q \)[/tex] is given by the Euclidean distance formula:
[tex]\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Where [tex]\( P(5, -3) \)[/tex] and [tex]\( Q(0, 1) \)[/tex].
Plug in the coordinates of [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]:
[tex]\[ PQ = \sqrt{(0 - 5)^2 + (1 - (-3))^2} \][/tex]
[tex]\[ PQ = \sqrt{(-5)^2 + (1 + 3)^2} \][/tex]
[tex]\[ PQ = \sqrt{25 + 16} \][/tex]
[tex]\[ PQ = \sqrt{41} \][/tex]
2. Calculate the distance [tex]\( QR \)[/tex]:
The distance [tex]\( QR \)[/tex] between points [tex]\( Q \)[/tex] and [tex]\( R \)[/tex] is also given by the Euclidean distance formula. Where [tex]\( Q(0, 1) \)[/tex] and [tex]\( R(x, 6) \)[/tex].
[tex]\[ QR = \sqrt{(x - 0)^2 + (6 - 1)^2} \][/tex]
[tex]\[ QR = \sqrt{x^2 + 5^2} \][/tex]
[tex]\[ QR = \sqrt{x^2 + 25} \][/tex]
3. Set the distances [tex]\( PQ \)[/tex] and [tex]\( QR \)[/tex] equal:
Since [tex]\( PQ = QR \)[/tex], we set the equations equal to each other:
[tex]\[ \sqrt{41} = \sqrt{x^2 + 25} \][/tex]
4. Solve for [tex]\( x \)[/tex]:
Square both sides to eliminate the square roots:
[tex]\[ (\sqrt{41})^2 = (\sqrt{x^2 + 25})^2 \][/tex]
[tex]\[ 41 = x^2 + 25 \][/tex]
Subtract 25 from both sides to isolate [tex]\( x^2 \)[/tex]:
[tex]\[ 41 - 25 = x^2 \][/tex]
[tex]\[ 16 = x^2 \][/tex]
Solve for [tex]\( x \)[/tex] by taking the square root of both sides:
[tex]\[ x = \pm \sqrt{16} \][/tex]
[tex]\[ x = \pm 4 \][/tex]
5. Verify the solution:
So, the solutions for [tex]\( x \)[/tex] are [tex]\( x = 4 \)[/tex] and [tex]\( x = -4 \)[/tex].
To ensure both solutions are valid, we can substitute them back into the distances [tex]\( PQ \)[/tex] and [tex]\( QR \)[/tex] if needed. Both [tex]\( x = 4 \)[/tex] and [tex]\( x = -4 \)[/tex] provide the correct distances that satisfy the initial condition [tex]\( PQ = QR \)[/tex].
Final Answer:
The values of [tex]\( x \)[/tex] that satisfy the condition where the distance [tex]\( PQ \)[/tex] is equal to the distance [tex]\( QR \)[/tex] are:
[tex]\[ x = 4 \quad \text{and} \quad x = -4 \][/tex]
Problem: Find [tex]\( x \)[/tex] such that the distance between points [tex]\( P(5, -3) \)[/tex] and [tex]\( Q(0, 1) \)[/tex] is equal to the distance between points [tex]\( Q(0, 1) \)[/tex] and [tex]\( R(x, 6) \)[/tex].
Step-by-Step Solution:
1. Calculate the distance [tex]\( PQ \)[/tex]:
The distance [tex]\( PQ \)[/tex] between points [tex]\( P \)[/tex] and [tex]\( Q \)[/tex] is given by the Euclidean distance formula:
[tex]\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Where [tex]\( P(5, -3) \)[/tex] and [tex]\( Q(0, 1) \)[/tex].
Plug in the coordinates of [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]:
[tex]\[ PQ = \sqrt{(0 - 5)^2 + (1 - (-3))^2} \][/tex]
[tex]\[ PQ = \sqrt{(-5)^2 + (1 + 3)^2} \][/tex]
[tex]\[ PQ = \sqrt{25 + 16} \][/tex]
[tex]\[ PQ = \sqrt{41} \][/tex]
2. Calculate the distance [tex]\( QR \)[/tex]:
The distance [tex]\( QR \)[/tex] between points [tex]\( Q \)[/tex] and [tex]\( R \)[/tex] is also given by the Euclidean distance formula. Where [tex]\( Q(0, 1) \)[/tex] and [tex]\( R(x, 6) \)[/tex].
[tex]\[ QR = \sqrt{(x - 0)^2 + (6 - 1)^2} \][/tex]
[tex]\[ QR = \sqrt{x^2 + 5^2} \][/tex]
[tex]\[ QR = \sqrt{x^2 + 25} \][/tex]
3. Set the distances [tex]\( PQ \)[/tex] and [tex]\( QR \)[/tex] equal:
Since [tex]\( PQ = QR \)[/tex], we set the equations equal to each other:
[tex]\[ \sqrt{41} = \sqrt{x^2 + 25} \][/tex]
4. Solve for [tex]\( x \)[/tex]:
Square both sides to eliminate the square roots:
[tex]\[ (\sqrt{41})^2 = (\sqrt{x^2 + 25})^2 \][/tex]
[tex]\[ 41 = x^2 + 25 \][/tex]
Subtract 25 from both sides to isolate [tex]\( x^2 \)[/tex]:
[tex]\[ 41 - 25 = x^2 \][/tex]
[tex]\[ 16 = x^2 \][/tex]
Solve for [tex]\( x \)[/tex] by taking the square root of both sides:
[tex]\[ x = \pm \sqrt{16} \][/tex]
[tex]\[ x = \pm 4 \][/tex]
5. Verify the solution:
So, the solutions for [tex]\( x \)[/tex] are [tex]\( x = 4 \)[/tex] and [tex]\( x = -4 \)[/tex].
To ensure both solutions are valid, we can substitute them back into the distances [tex]\( PQ \)[/tex] and [tex]\( QR \)[/tex] if needed. Both [tex]\( x = 4 \)[/tex] and [tex]\( x = -4 \)[/tex] provide the correct distances that satisfy the initial condition [tex]\( PQ = QR \)[/tex].
Final Answer:
The values of [tex]\( x \)[/tex] that satisfy the condition where the distance [tex]\( PQ \)[/tex] is equal to the distance [tex]\( QR \)[/tex] are:
[tex]\[ x = 4 \quad \text{and} \quad x = -4 \][/tex]
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