Question

Find the relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3 , 5)​

Answer 1

Answer:

x-y-2=0

Step-by-step explanation:

Let the points be A and B.

Let P be the point equidistant from A and B.

A=(7,1), B=(3,5), P=(x,y)

Distance between AP=Distance between PB

√(x1-x2)²+(y1-y2)²=√(x1-x2)²+(y1-y2)²

√(7-x)²+(1-y)²=√(x-3)²+(y-5)²

Squaring on both the sides

(√(7-x)²+(1-y)²)²=(√(x-3)²+(y-5)²)²

(7-x)²+(1-y)²=(x-3)²+(y-5)²

49-14x+x²+1-2y+y²=x²-6x+9+y²-10y+25

x²-x²+y²-y²-14x+6x-2y+10y+49+1-9-25=0

-8x+8y+16=0

-x+y+2=0

x-y-2=0

Therefore, the relation between x and y is x-y-2=0.

Answer 2

❍ Given that, point C(x, y) is equidistant from the points A(7, 1) and B(3, 5).

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As we know that,

To calculate the distance b/w two given points (x₁, x₂) and (y₁, y₂) formula is given by :

$\bigstar\:{\underline{\boxed{\sf{\sqrt{\bigg(x_2 - x_1 \bigg)^2 + \bigg(y_2 - y_1 \bigg)^2}}}}}$

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$\dashrightarrow\sf \sqrt{\Big(7 - x\Big)^2 + \Big(1 - y\Big)^2} = \sqrt{\Big(3 - x\Big)^2 + \Big(5 - y^2\Big)^2 }\\\\\\\dashrightarrow\sf \Bigg(\sqrt{\Big(7 - x\Big)^2 + \Big(1 - y\Big)^2}\Bigg)^2 = \Bigg(\sqrt{\Big(3 - x\Big)^2 + \Big(5 - y\Big)^2}\Bigg)^2\\\\\\\dashrightarrow\sf \Big(7 - x\Big)^2 + \Big(1 - y\Big)^2 = \Big(3 - x\Big)^2 = \Big(5 - y\Big)^2\\\\\\\dashrightarrow\sf 7^2 +x^2 - 14x + 1^2 + y^2 -2y = 3^2 + x^2 - 6x + 5^2 + y^2 - 10y\\\\\\\dashrightarrow\sf 49 + x^2 - 14x + 1+ y^2 - 2y = 9 + x^2 - 6x + 25 - y^2 - 10y\\\\\\\dashrightarrow\sf x^2 - x^2 - 14x + 6x + y^2 - y^2 - 2y + 10y + 9 - 49 - 1 + 25 = 0\\\\\\\dashrightarrow\sf -8x + 8y + 16 = 0\\\\\\\dashrightarrow\sf 8\Big\{-x + y + 2\Big\} = 0\\\\\\\dashrightarrow\sf -x + y + 2 = \cancel\dfrac{0}{8}\\\\\\\dashrightarrow\sf y + 2 = x\\\\\\\dashrightarrow\underline{\boxed{\pmb{\frak{\pink{x = y + 2}}}}}\;\bigstar$

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$\therefore\:{\underline{\textsf{Hence,\:the\:relation\:between\:x\:and\:y\:is\: {\textbf{(x = y + 2)}}.}}}$