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Q.

The point $(x, y)$ is equidistant from (7,1) and (3,5), such that x – y = ____.

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Detailed Solution

The point (x, y) is equidistant from (7,1) and (3,5), such that x – y = 2.
We have been given,
The point (x, y) is equidistant from (7,1) and (3,5).
Using the distance formula,

D = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}

Using the identity,

{(a - b)}^{2} = a^{2} - 2 ab + b^{2}

The distance between the points (x,y) and (7,1) is therefore equivalent to the distance between (3,5).
Now the distance between the points (x,y) and (7,1),
Substituting

x_{2} = 7, x_{1} = x, y_{2} = 1, and y_{1} = y

, in the formula,

\Rightarrow \sqrt{{(7 - x)}^{2} + {(1 - y)}^{2}}

And the distance between points,

(x, y)

and

(3, 5)

,
Substituting

x_{2} = 3, x_{1} = x, y_{2} = 5, and y_{1} = y

, in the formula,

\Rightarrow \sqrt{{(3 - x)}^{2} + {(5 - y)}^{2}}

As the distance are equal, therefore, equating both the equations,

(7 − x) 2 + (1 − y) 2 = (3 − x) 2 + (5 − y) 2 ⇒ 49 − 14 x + x 2 + 1 − 2 y + y 2 = 9 − 6 x + x 2 + 25 − 10 y + y 2 ⇒ 8 x − 8 y + 34 − 50 = 0 ⇒ x − y = 2

Therefore, the correct answer is 2.

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