Question

Q7. Find the relation between x and y such that the point (x, y) is equidistant from the point (3,6) and (-3, 4).​

Answer 1

Question:

• Find the relation between x and y such that the point (x, y) is equidistant from the point (3,6) and (-3, 4).

Answer:

• Relation b/w x & y is 3x + y = 5.

$\quad{\huge{\boxed{\textsf{\textbf{\red{E}\green{X}\purple{P}\pink{L}\blue{A}\orange{N}A\gray{T}\red{I}\green{0}\purple{N}}}}}}$

Given that:

• Point (x, y) is equidistant from the point (3, 6) and (-3, 4).

To Find:

• Relation between x and y?

Solution:

• Identity to know before solving this question ::

$\boxed{\sf{\red{(a - b)^2 = a^2 + b^2 - 2ab}}}$

Formula used:

$\boxed{\sf{\purple{Distance\:Formula = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}\:\:}}}$

$\:$

• Finding distance b/w (x, y) & (3, 6) ::

By using distance formula we get,

$\\ :\implies\footnotesize\sf \sqrt{(3 - x)^2 + (6 - y)^2}$

$\\ :\implies\footnotesize\sf \sqrt{(3)^2 + (x)^2 - (2\:\times\:3\:\times\:x) + (6)^2 + (y)^2 - (2\:\times\:6\:\times\:y)}$

$\\ :\implies\footnotesize\sf \sqrt{9 + x^2 - 6x + 36 + y^2 - 12y}$

$\\ :\implies\boxed{\bf{\sqrt{9 + x^2 - 6x + 36 + y^2 - 12y}}}$

$\:$

• Finding distance b/w (x, y) & (3, 6) ::

By using distance formula we get,

$\\ :\implies\footnotesize\sf \sqrt{(-3 - x)^2 + (4 - y)^2}$

$\\ :\implies\footnotesize\sf \sqrt{(-3)^2 + (x)^2 - [2\:\times\:(-3)\:\times\:x] + (4)^2 + (y)^2 - (2\:\times\:4\:\times\:y)}$

$\\ :\implies\footnotesize\sf \sqrt{9 + x^2 - (-6x) + 16 + y^2 - 8y}$

$\\ :\implies\boxed{\bf{\sqrt{9 + x^2 + 6x + 16 + y^2 - 8y}}}$

$\:$

Now,

According to the question,

• Point (x, y) is equidistant from the point (3,6) & (-3, 4).

Therefore,

$\\ :\implies\footnotesize\sf\sqrt{9 + x^2 - 6x + 36 + y^2 - 12y} = \sqrt{9 + x^2 + 6x + 16 + y^2 - 8y}$

On squaring both sides we get,

$\\ :\implies\footnotesize\sf (\sqrt{9 + x^2 - 6x + 36 + y^2 - 12y})^2 = (\sqrt{9 + x^2 + 6x + 16 + y^2 - 8y})^2$

$\\ :\implies\footnotesize\sf \cancel{9} + \cancel{x^2} - 6x + 36 + \cancel{y^2} - 12y = \cancel{9} + \cancel{x^2} + 6x + 16 + \cancel{y^2} - 8y$

$\\ :\implies\footnotesize\sf -6x + 36 - 12y = 6x + 16 - 8y$

$\\ :\implies\footnotesize\sf 36 - 16 = 6x - 8y + 6x + 12y$

$\\ :\implies\footnotesize\sf 6x + 6x - 8y + 12y = 20$

$\\ :\implies\footnotesize\sf 12x + 4y = 20$

$\\ :\implies\footnotesize\sf 4(3x + y) = 20$

$\\ :\implies\footnotesize\sf 3x + y = {\cancel{\dfrac{20}{4}}}$

$\\ :\implies\boxed{\bf{3x + y = 5}}$

∴ Relation b/w x & y is $\large{\boxed{\bf{\red{3}\green{x} \purple{+} \pink{y} \blue{=} \orange{5}}}}$

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Answer 2

$\pink{ \boxed{\boxed{\begin{array}{cc}\bf \: The \: relation \: between \: x \: an d \: y \: is \: : \\ \\ \bf \: 3x + y = 5.\end{array}}}}$

Given,

The point (x, y) is equidistant from the point (3,6) and (-3, 4).

To find :

Relation between x and y

Solution :

Let,

→ P(x,y)

→ Q(3,6)

→ R(-3,4)

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

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

${ \boxed{\boxed{\begin{array}{cc} \rm \: distance \: between \: P \: \: and \: \: R \: is \: : \: \\ \\ \rm \: PR = \sqrt{ {(x + 3)}^{2} + {(y - 4)}^{2} } \end{array}}}}$

and

${ \boxed{\boxed{\begin{array}{cc} \rm \: distance \: between \: \: P \: \: and \: \: Q \: is \: : \\ \\ PQ = \sqrt{ {(x - 3)}^{2} + {(y - 6)}^{2} } \end{array}}}}$

According to the question,

$\small{ \boxed{ \begin{array}{cc}PQ = PR \\ \\ \rm \implies \: \sqrt{ {(x - 3)}^{2} + {(y - 6)}^{2} } = \sqrt{ {(x + 3)}^{2} + {(y - 4)}^{2} } \\ \\ \rm \implies \: { {(x - 3)}^{2} + {(y - 6)}^{2} } = { {(x + 3)}^{2} + {(y - 4)}^{2} } \\ \\ \rm \implies \: {x}^{2} + 9 - 6x + {y}^{2} + 36 - 12y = {x}^{2} + 9 + 6x + {y}^{2} + 16 - 8y \\ \\ \rm \implies \: - 6x + 45 - 12y = 6x - 8y + 25 \\ \\ \rm \implies \: - 6x - 6x - 12y + 8y = 25 - 45 \\ \\ \rm \implies \: - 12x - 4y = - 20 \\ \\ \rm \implies \: 3x + y = 5 \end{array}}}$

So,

The relation between x and y is = 3x + y = 5