Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).
Answers
Step-by-step explanation:
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Given : Point (x, y) is equidistant from the points (3, 6) and (-3, 4).
Let P(x , y) , A( 3, 6) and B( - 3 , 4)
AP = BP [equidistant from P]
√{(x - 3)² + (y - 6)² } = √{(x + 3)² + (y - 4)²}
[Distance Formula =√(x2 - x1)² + (y2 - y1)²]
On squaring both sides
{(x - 3)² + (y - 6)² } = √{(x + 3)² + (y - 4)²
[x² + 3² - 2 × x × 3 + y² + 6² - 2 × x × 6] = [x² + 3² + 2 × x × 3 + y² + 4² - 2 × x × 4]
[(a - b)² = a² + b² - 2ab]
[x² + 9 - 6x + y² + 36 - 12y ] = [x² + 9 + 6x + y² + 16 - 8y]
[x² - 6x + y² + 45 - 12y ] = [x² + 6x + y² + 25 - 8y]
x² - x² + y² - y² - 6x - 6x - 12y - 8y = 25 - 45
-12 x - 20y = - 20
-4(3x + 5y) = - 20
(3x + 5y) = - 20/4
(3x + 5y) = 5
(3x + 5y) - 5 = 0
Hence , a relation between x and y is (3x + 5y) - 5 = 0
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