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Find a relation between x and y such that point$ (x, y)$ is equidistant from the point (7, 1) and (3, 5)
Given: The points are $ (7, 1) \ and \ (3, 5)$
To Do: Find a relation between x and y such that point $ (x, y)$ is equidistant from the given points
Solution:
Formula to find distance between two points
$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Distance between $ (x , y) \ and \ (7 , 1)$
${x_1 = x ; x_2 = 7 ; y_1 = y ; y_2 = 1}$
substitute in formula
$\sqrt{(7 - x)^2 + (1 - y)^2}$ [ $(a-b)^2 = a^2 - 2 a b + b^2$ ]
$\sqrt{49 - 14x + x^2 + 1 - 2 y + y^2}$ .......................................................(i)
Distance between (x , y) and (3 , 5)
$x_1 = x ; x_2 = 3 ; y_1 = y ; y_2 = 5$
substitute in formula
$\sqrt{(3 - x)^2 + (5 - y)^2}$ [ $(a-b)^2 = a2 - 2 a b + b^2$ ]
$\sqrt{9 - 6 x + x^ 2 + 25 - 10 y + y^2}$ .......................................................(ii)
Since, two distance are equal, equate (i) and (ii)
$\sqrt{49 - 14x + x^2 + 1 - 2 y + y^2}$ = $\sqrt{9 - 6 x + x^ 2 + 25 - 10 y + y^ 2}$
Square roots on both sides, so they get cancel
$49 - 14x + x^2 + 1 - 2 y + y^ 2 = 9 - 6 x + x^ 2 + 25 - 10 y + y^ 2$
$49 - 14x + x^2 + 1 - 2 y + y^ 2 - 9 + 6 x - x^ 2 - 25 + 10 y - y^ 2 = 0$
$x2 - x^2 + y^ 2 + y^ 2 -14 x + 6 x - 2 y + 10 y + 49 + 1 - 9 - 25 = 0$
$-14 x + 6 x - 2 y + 10 y + 50 - 34 = 0 $ [$ x^2 - x^2 = 0 ; y^ 2 + y^ 2 = 0$]
$-8 x + 8 y + 16 = 0$
divide by 8,
$- x + y + 2 = 0$
$y - x = -2$
Therefore the relation between $x$ and $y$ is $y - x = -2$
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