To find a relation between x and y such that the point P(x,y) is equidistant from the points A(-5,3) and B(7,2), we can use the distance formula. The distance formula is given by:

d = sqrt[(x2 - x1)^2 + (y2 - y1)^2]

We know that the distance from P to A is equal to the distance from P to B. Therefore, we can set up the following equation:

sqrt[(x - (-5))^2 + (y - 3)^2] = sqrt[(x - 7)^2 + (y - 2)^2]

Squaring both sides to get rid of the square root gives us:

(x + 5)^2 + (y - 3)^2 = (x - 7)^2 + (y - 2)^2

Expanding and simplifying gives us:

x^2 + 10x + 25 + y^2 - 6y + 9 = x^2 - 14x + 49 + y^2 - 4y + 4

Subtracting x^2, y^2, 10x, and -6y from both sides gives us:

24x + 2y = 19

So, the relation between x and y such that the point P(x,y) is equidistant from the points A(-5,3) and B(7,2) is 24x + 2y = 19.

Question

To find a relation between x and y such that the point P(x,y) is equidistant from the points A(-5,3) and B(7,2), we can use the distance formula. The distance formula is given by:

d = sqrt[(x2 - x1)^2 + (y2 - y1)^2]

We know that the distance from P to A is equal to the distance from P to B. Therefore, we can set up the following equation:

sqrt[(x - (-5))^2 + (y - 3)^2] = sqrt[(x - 7)^2 + (y - 2)^2]

Squaring both sides to get rid of the square root gives us:

(x + 5)^2 + (y - 3)^2 = (x - 7)^2 + (y - 2)^2

Expanding and simplifying gives us:

x^2 + 10x + 25 + y^2 - 6y + 9 = x^2 - 14x + 49 + y^2 - 4y + 4

Subtracting x^2, y^2, 10x, and -6y from both sides gives us:

24x + 2y = 19

So, the relation between x and y such that the point P(x,y) is equidistant from the points A(-5,3) and B(7,2) is 24x + 2y = 19.

Knowee AI · Accepted Answer

To find a relation between x and y such that the point P(x,y) is equidistant from the points A(-5,3) and B(7,2), we can use the distance formula. The distance formula is given by:

d = sqrt[(x2 - x1)^2 + (y2 - y1)^2]

We know that the distance from P to A is equal to the distance from P to B. Therefore, we can set up the following equation:

sqrt[(x - (-5))^2 + (y - 3)^2] = sqrt[(x - 7)^2 + (y - 2)^2]

Squaring both sides to get rid of the square root gives us:

(x + 5)^2 + (y - 3)^2 = (x - 7)^2 + (y - 2)^2

Expanding and simplifying gives us:

x^2 + 10x + 25 + y^2 - 6y + 9 = x^2 - 14x + 49 + y^2 - 4y + 4

Subtracting x^2, y^2, 10x, and -6y from both sides gives us:

24x + 2y = 19

So, the relation between x and y such that the point P(x,y) is equidistant from the points A(-5,3) and B(7,2) is 24x + 2y = 19.

Find a relation between x and y such that the point , P(x,y) is equidistant from the points , A (-5,3) and ,B(7,2).

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