Question

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



x+y = 2

x-y + 2

x+y = 3

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Answer 1

Answer:

Good morning Siso

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Answer 2

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 :-

$\bigg(\dfrac{7 + 3}{2} \: , \: \dfrac{1 + 5}{2} \bigg)⟹( 27+3, 21+5 )$

$\rm\implies \bigg( \dfrac{10}{2} \: , \: \dfrac{6}{2} \bigg)⟹( 210 , 26 )</p><p>$

$</p><p>\rm \implies \bigg( 5 \: , \:3\bigg)⟹(5,3)</p><p>$

$</p><p>\therefore \rm(x,y) = (5,3)∴(x,y)=(5,3)$

=> 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 :-

$\rm \implies \bigg( \dfrac{a + c}{2} \: , \: \dfrac{b + d}{2} \bigg)⟹( 2a+c , 2b+d )</p><p>$

I hope it is helpful for you