Question

if x-y=2 then point (x,y) is equidistant from (7,1) and (__,__)​

Answer 1

The point is $\left(\dfrac{15}{2},\dfrac{1}{2}\right)$

Step-by-step explanation:

If x - y = 2

P(x,y) is equidistant from A(7,1) and B(a,b)

Therefore, PA=PB

$\sqrt{(x-7)^2+(y-1)^1}=\sqrt{(x-a)^2+(y-b)^2}$

$x^2+49-14x+y^2+1-2y=x^2+a^2-2xa+y^2+b^2-2yb$

$-14x+2xa-2y+2yb=a^2+b^2-50$

$x(2a-14)-y(2-2b)=a^2+b^2-50$

Now compare the equation with x-y=2

So, 2a-14=1   , 2-2b=1   , $a^2+b^2-50=2$

$a=\dfrac{15}{2},b=\dfrac{1}{2}$

The point is $\left(\dfrac{15}{2},\dfrac{1}{2}\right)$

#BAL

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)

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