Question

find the relation between x and y if the point A(x,y) is equidistant form the point (3,7) and (5,4)​

Answer 1

Answer:

Let P(x,y) be equidistant from the points A(7,1) and B(3,5).

AP=BP

AP

2

=BP

2

(x−7)

2

+(y−1)

2

=(x−3)

2

+(y−5)

2

x

2

−14x+49+y

2

−2y+1=x

2

−6x+9+y

2

−10y+25

x−y=2

Step-by-step explanation:

Hope it helps you

Answer 2

[tex]\mathbb{ \bold{ \huge{Answer:- } } } \\Let \: P \: be \: the \: point \: equidistant \: from \: AB \\ A=(3,7), P=(x, y), B=(5,4) \\ Distance \: between \: AP= Distance \: between \: BP \\ \sqrt{ {(x_{2} - x_{1}}^{2} ) + {(y_{2} - y_{1})}^{2} } = \sqrt{ {(x_{2} - x_{1}}^{2} ) + {(y_{2} - y_{1})}^{2} } \\ \sqrt{ {(x - 3)}^{2} + {(y - 7)}^{2}} = \sqrt{ {(x - 5)}^{2} + {(y - 4)}^{2} } \\Squaring \: on \: both \: the \: sides \\ {(\sqrt{ {(x - 3)}^{2} + {(y - 7)}^{2}})}^{2} = {( \sqrt{ {(x - 5)}^{2} + {(y - 4)}^{2} } )}^{2} \\ {(x - 3)}^{2} + {(y - 7)}^{2} = {(x - 5)}^{2} + {(y - 4)}^{2} \\ {x}^{2} + 9 - 6x + {y}^{2} + 49 - 14y = {x}^{2} + 25 - 10x + {y}^{2} + 16 - 8y \\ {x}^{2} - {x}^{2} + {y}^{2} - {y}^{2} - 6x + 10x - 14y + 8y +9 + 49 - 25 - 16 = 0 \\ \mathbb{ \bold{ \large{\boxed{{4x - 6y + 17 = 0 } } } }}[/tex]