The Point (x,y) is Equidistant from the Point is (7,1) and (3,5) Then: (a) x+y=2 (b) -x+y=2 (c) x-y=2 (d) -x-y=2 - Bzziii

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The point (`x,y`) is equidistant from the poin ts (7,1) and (3,5) Then:

(a) x+y=2
(b) -x+y=2
(c) x-y=2
(d) -x-y=2





(c) x-y=2



Option (c)x-y=2 is the correct answer.

Here, 

(x, y) ⇒ `x_1` =  x ,   `y_1` = y

(7, 1) ⇒ `x_2` =  7 ,   `y_2` = 1

(3, 5) ⇒ `x_3` =  7 ,   `y_3` = 5

A/c , `\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}` = `\sqrt{(x_{3}-x_{1})^2+(y_{3}-y_{1})^2}`

⇒ `\sqrt{(7 -"x")^2+(1-"y")^2}` = `\sqrt{(3-"x")^2+(5-"y")}`

⇒ `7^2 - 2.7."x" + "x"^2 + 1^2 - 2.1.y + y^2` = `3^2 - 2.3."x" + "x"^2 + 5^2 - 2.5.y + y^2`

⇒ 49 - 14x + `"x"^2` +1 - 2y + `"y"^2` = 9 - 6x + `"x"^2` +25 + -10y + `"y"^2`

⇒ -14x + 6x - 2y + 10y = 34 - 50

⇒ -8x + 8y = -16

⇒  -8 (x - y) = -16

⇒ x - y = `-16/-8`

⇒  x - y = 2

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