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 :

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

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

and

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

According to the question,

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

So,

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

Answer 2

Answer:

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

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} } \: \: unitd=(x1−x2)2+(y1−y2)2unit

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

and

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

According to the question,

\begin{gathered}\small{ \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}}}\end{gathered}PQ=PR⟹(x−2)2+(y−1)2=(x+2)2+(y−3)2⟹(x−2)2+(y−1)2=(x+2)2+(y−3)2⟹x2−4x+4+y2−2y+1=x2+4x+4+y2−6y+9⟹−4x−4x−2y+6y=4+9−4−1⟹−8x+4y=8⟹−2x+y=2⟹2x−y+2=0

So,

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