Question

find the relation between such that the x,y is equidistant (2,1) and (-2,3)​

Answer 1

$\pink{ \boxed{\boxed{\begin{array}{cc}\bf \: The \: relation \: between \: x \: an d \: y \: is \: : \\ \\ \bf \: 2x - y+2= 0.\end{array}}}}$

Given,

The point (x, y) is equidistant from the point (2,1) and (-2, 3).

To find :

Relation between x and y

Solution :

Let,

→ P(x,y)

→ Q(2,1)

→ R(-2,3)

we know that, the formula of distance between two point is :

$\rm \: d = \sqrt{ {(x_1 - x_2)}^{2} + {(y_1 - y_2)}^{2} } \: \: unit$

${ \pink{\boxed{\boxed{\begin{array}{cc} \rm \: distance \: between \: P \: \: and \: \: R \: is \: : \: \\ \\ \rm \: PR = \sqrt{ {(x + 2)}^{2} + {(y - 3)}^{2} } \end{array}}}}}$

and

${ \blue{\boxed{\boxed{\begin{array}{cc} \rm \: distance \: between \: \: P \: \: and \: \: Q \: is \: : \\ \\ PQ = \sqrt{ {(x - 2)}^{2} + {(y - 1)}^{2} } \end{array}}}}}$

According to the question,

$\small{\orange{ \boxed{ \begin{array}{cc}PQ = PR \\ \\ \rm \implies \: \sqrt{ {(x - 2)}^{2} + {(y - 1)}^{2} } = \sqrt{ {(x + 2)}^{2} + {(y - 3)}^{2} } \\ \\ \rm \implies \: { {(x - 2)}^{2} + {(y - 1)}^{2} } = { {(x + 2)}^{2} + {(y - 3)}^{2} } \\ \\ \rm \implies \: {x}^{2} - 4x +4 + {y}^{2} - 2y + 1 = {x}^{2} + 4x + 4 + {y}^{2} - 6y + 9 \\ \\ \rm \implies \: - 4x - 4x - 2y + 6y = 4 + 9 - 4 - 1 \\ \\ \rm \implies \: - 8x + 4y = 8 \\ \\ \rm \implies \: - 2x + y = 2 \\ \\ \rm \implies \:2x - y + 2 = 0 \end{array}}}}$

So,

The relation between x and y is : 2x - y+2 = 0

Answer 2

Answer:

Coordinates of the points are A(3, 6) and B(-3, 4) Let the point P(x, y) is equidistant from A and B

Using distance formula = √ ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 (x2−x1)2+(y2−y1)2 ⇒ PA = PB √ ( x − 3 ) 2 + ( y − 6 ) 2 (x−3)2+(y−6)2 = √ ( x + 3 ) 2 + ( y − 4 ) 2 (x+3)2+(y−4)2 On squaring both sides, we get (x - 3)2 + (y - 6)2 = (x + 3)2 + (y - 4)2 x2 - 6x + 9 + y2 - 12y + 36 = x2 + 6x + 9 + y2 - 8y + 16 3x + y = 5