Question

find the relationship between x and y such that the point xy is equidistant from the point (17,-9) (-3,11)​

Answer 1

Given:

A point (x,y) is equidistant from the point (17,-9) (-3,11)​.

TO find:

The relation between x and y

Solution:

Let the point M is (x,y)

Here we will use the distance formula  to establish the relation:

$\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

as we have

A(17,-9) and B(-3,11)

according to the question:

AM=BM

So,  we get

$\sqrt{(17-x)^{2}+(-9-y)^{2}} = \sqrt{(-3-x)^{2}+(11-y)^{2}}$

Squaring both side

• $(17-x)^{2}+(-9-y)^{2}=(-3-x)^{2}+(11-y)^{2}$
• 289+x²-34x + 81+y²+18y = 9+x²+6x+121+y²-22y
• 240 = 40x-40y
• x-y = 240/40
• x-y = 6

So, the relation between x and y is x-y = 6.

Answer 2

Step-by-step explanation:

Given:Point (17,-9) (-3,11)

To find:Find the relationship between x and y such that the point P(x,y) is equidistant

Solution:

Distance formula

$\sqrt{( {x_2 - x_1)}^{2} + ( {y_2 - y_1)}^{2} }$

A(17,-9) and B(-3,11)

AP=BP

So,

$AP^2=BP^2\\\\ \sqrt{( {17 - x)}^{2} + ( { - 9 - y)}^{2} } = \sqrt{( { - 3 - x)}^{2} + ( { 11 - y)}^{2} }$

Square both side

$( {17 - x)}^{2} + ( { - 9 - y)}^{2} = ( { - 3 - x)}^{2} + ( { 11 - y)}^{2} \\ \\ 289 + {x}^{2} - 34x + 81 + {y}^{2} + 18y = 9 + {x}^{2} + 6x + 121 + {y}^{2} - 22y \\ \\ cancel \: common \: term \\ \\ - 34x - 6x + 18y + 22y + 289 + 81 - 9 - 121 = 0 \\ \\ - 40x + 40y + 240 = 0 \\ \\ \bold{\pink{- x + y + 60 = 0 }}\\ \\ or \\ \\ \bold{\green{x - y = 60}}$

So,the relation between x and y is

x-y=60 or -x+y+60=0.

Hope it helps you.

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