Question

If x-y=2 then point (x,y) is equidistant from (7,1) and .......
The answer is (3,5) ,but how do you solve it ??​

Answer 1

Step-by-step explanation:

x=2+y

Let (x,y) be equidistant from (7,1) and (a,b)

2+y=7+a/2

y=1+b/2

7+a-4/2=1+b/2

3+a=1+b

a+2=b

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Answer 2

If x-y=2 then point (x,y) is equidistant from (7,1) and (3,5)

Step-by-step explanation:

if x-y = 2..(1)

point (x,y) is equidistant from (7,1) and (a,b)

Then a,b will be the image of 7,1

x = $\frac{7+a}{2}$

2x = 7+a

$y = \frac{1+b}{2}\\ 2y = 1+b$

then on solving (putting the value of y from 1)

2(x-2)= 1+b

2x -4= 1+b

2x+ b+5

now putting the value of x

7+a = b+5

b = a+2

slope of x-y = 2

y = x-2 is 1

therefore , slope of (a,b) and (7,1) is -1

then

$\frac{b-1}{a-7}= -1\\ b = 8-a$

now , a+2 = 8-a

a = 3

b= 5

hence,

If x-y=2 then point (x,y) is equidistant from (7,1) and (3,5)

#Learn more:

If (x,y) is equidistant from points (7,1) and (3,5),show that y=x-2

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