Question

A relation between x and y such that the pt. (x,y) is equidistant from the points (7,1) and (3,5)

2 points

x+y = 2

x-y + 2

x+y = 3


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Answer 1

Correct question :-

A relation between x and y such that the pt. (x,y) is equidistant from the points (7,1) and (3,5) :-

a) x+y = 2

b) x-y = 2

c) x+y = 3

Solution :-

If point (x ,y) is equidistant from the points (7,1) and (3,5). So (x,y) is midpoint of line joining (7,1) and (3,5).

Co-ordinates of midpoint of line joining (7,1) and (3,5) can be written as :-

$\rm \implies \bigg( \dfrac{7 + 3}{2} \: , \: \dfrac{1 + 5}{2} \bigg)$

$\rm \implies \bigg( \dfrac{10}{2} \: , \: \dfrac{6}{2} \bigg)$

$\rm \implies \bigg( 5 \: , \:3\bigg)$

$\therefore \rm(x,y) = (5,3)$

=> x = 5

=> y = 3

Now , x - y = 5-3 = 2

Therefore option (b) is correct.

Answer :

• b) x-y = 2

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Learn more :

Co-ordinates of midpoint of line joining (a,b) and (c,d) is given as :-

$\rm \implies \bigg( \dfrac{a + c}{2} \: , \: \dfrac{b + d}{2} \bigg)$

Answer 2

⭐Answer⭐

➡ Your answer : x-y=2

☆ Solution :-

Let P (x,y) be equidistant from the points A(7,1) and B(3,5).

AP = BP

AP ^2 = BP ^2

=> (x − 7) ^2 +(y − 1) ^2 =(x − 3) ^2+

(y − 5)^2

=> x^2 − 14x + 49 + y^2 − 2y + 1= x^2 − 6x+ 9 + y^2 − 10y + 25

=> x − y = 2

∴Option (b) x - y = 2 is the correct answer

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