Question
If a p (X ,Y )is equal distance to from a (5, 1 )and b (- 1,5 )then 3 X is equals to
Answers
Answered by
oobleck
If P=(x,y) is the midpoint of AB, then
x = (5+1)/2 = 3
so 3x = 9
of course, as originally stated, P could be anywhere on the perpendicular bisector of AB.
x = (5+1)/2 = 3
so 3x = 9
of course, as originally stated, P could be anywhere on the perpendicular bisector of AB.
Answered by
GPT-4o
To solve the problem of finding the value of \(3X\) given that a point \( P(X, Y) \) is equidistant from points \( A(5, 1) \) and \( B(-1, 5) \), we need to set the distances from \( P \) to \( A \) and \( P \) to \( B \) as equal and solve for \( X \).
The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For point \( P(X, Y) \) and point \( A(5, 1) \), the distance is:
\[ PA = \sqrt{(X - 5)^2 + (Y - 1)^2} \]
For point \( P(X, Y) \) and point \( B(-1, 5) \), the distance is:
\[ PB = \sqrt{(X + 1)^2 + (Y - 5)^2} \]
Given that \( P \) is equidistant from \( A \) and \( B \):
\[ PA = PB \]
Thus,
\[ \sqrt{(X - 5)^2 + (Y - 1)^2} = \sqrt{(X + 1)^2 + (Y - 5)^2} \]
Square both sides to eliminate the square roots:
\[ (X - 5)^2 + (Y - 1)^2 = (X + 1)^2 + (Y - 5)^2 \]
Expand both sides:
\[ (X^2 - 10X + 25) + (Y^2 - 2Y + 1) = (X^2 + 2X + 1) + (Y^2 - 10Y + 25) \]
Combine like terms:
\[ X^2 - 10X + 25 + Y^2 - 2Y + 1 = X^2 + 2X + 1 + Y^2 - 10Y + 25 \]
\[ X^2 - 10X + 25 + Y^2 - 2Y + 1 - X^2 - 2X - 1 - Y^2 + 10Y - 25 = 0 \]
\[ -10X - 2X - 2Y + 10Y = 0 \]
Simplify:
\[ -12X + 8Y = 0 \]
Solve for \( X \):
\[ -12X = -8Y \]
\[ X = \frac{2}{3}Y \]
We need to find \( 3X \):
\[ 3X = 3 \left( \frac{2}{3}Y \right) = 2Y \]
Therefore, \( 3X = 2Y \).
The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For point \( P(X, Y) \) and point \( A(5, 1) \), the distance is:
\[ PA = \sqrt{(X - 5)^2 + (Y - 1)^2} \]
For point \( P(X, Y) \) and point \( B(-1, 5) \), the distance is:
\[ PB = \sqrt{(X + 1)^2 + (Y - 5)^2} \]
Given that \( P \) is equidistant from \( A \) and \( B \):
\[ PA = PB \]
Thus,
\[ \sqrt{(X - 5)^2 + (Y - 1)^2} = \sqrt{(X + 1)^2 + (Y - 5)^2} \]
Square both sides to eliminate the square roots:
\[ (X - 5)^2 + (Y - 1)^2 = (X + 1)^2 + (Y - 5)^2 \]
Expand both sides:
\[ (X^2 - 10X + 25) + (Y^2 - 2Y + 1) = (X^2 + 2X + 1) + (Y^2 - 10Y + 25) \]
Combine like terms:
\[ X^2 - 10X + 25 + Y^2 - 2Y + 1 = X^2 + 2X + 1 + Y^2 - 10Y + 25 \]
\[ X^2 - 10X + 25 + Y^2 - 2Y + 1 - X^2 - 2X - 1 - Y^2 + 10Y - 25 = 0 \]
\[ -10X - 2X - 2Y + 10Y = 0 \]
Simplify:
\[ -12X + 8Y = 0 \]
Solve for \( X \):
\[ -12X = -8Y \]
\[ X = \frac{2}{3}Y \]
We need to find \( 3X \):
\[ 3X = 3 \left( \frac{2}{3}Y \right) = 2Y \]
Therefore, \( 3X = 2Y \).
Answered by
GPT-4o
Given that point \( P(x, y) \) is the midpoint of \( A(5, 1) \) and \( B(-1, 5) \), we need to calculate the midpoint coordinates and subsequently determine \( 3x \).
The formula for the midpoint \( M \) of two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Plugging in the coordinates of points \( A(5, 1) \) and \( B(-1, 5) \):
\[ x = \frac{5 + (-1)}{2} = \frac{4}{2} = 2 \]
\[ y = \frac{1 + 5}{2} = \frac{6}{2} = 3 \]
Therefore, the coordinates of the midpoint \( P \) are:
\[ P(2, 3) \]
Now, to find \( 3x \):
\[ 3x = 3 \cdot 2 = 6 \]
So, \( 3x = 6 \).
The formula for the midpoint \( M \) of two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Plugging in the coordinates of points \( A(5, 1) \) and \( B(-1, 5) \):
\[ x = \frac{5 + (-1)}{2} = \frac{4}{2} = 2 \]
\[ y = \frac{1 + 5}{2} = \frac{6}{2} = 3 \]
Therefore, the coordinates of the midpoint \( P \) are:
\[ P(2, 3) \]
Now, to find \( 3x \):
\[ 3x = 3 \cdot 2 = 6 \]
So, \( 3x = 6 \).
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