Question

give two pairs of value of x and y such that(x,y) is equidistant from points (-1,8) and (3,4).

Answer 1

REFER THE ATTACHMENT.

Answer 2

The two pairs of value of x and y are (1, 6) and (2, 7).

Step-by-step explanation:

Let P(x, y) is equidistant from points A(- 1, 8) and B(3, 4).

To find, two pairs of value of x and y = ?

∴ $PA^{2} =PA^{2}$

Using distance formula,

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

${(x+1)^{2} +(y-8)^{2}}={(3-x)^{2} +(4-y)^{2}}$

⇒$x^{2}+2x+1 +y^{2}-16y+64= x^{2}-6x+9+ y^{2}-8y+16$

⇒$2x+1-16y+64=-6x+9-8y+16$

⇒ $8x-8y+40=0$

⇒$x-y+5=0$

⇒ $x=y-5$

Put y = 6, 7, we get

x = 1 , 2

Hence, the two pairs of value of x and y are (1, 6) and (2, 7).