Question

Find a relatiom between x and y such that the point p(x, y) is eqidistant from the point a(7, 0) and b(0, 5)​

Answer 1

❍ Given that, P(x, y) is eqidistant from the point A(7, 0) and B(0, 5).

Therefore,

$:\implies\sf AP = BP\\\\\\ :\implies\sf (AP)^2 = (BP)^2$

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As we know that, To find distance between two points x₁ , x₂ and y₁ , y₂ we use :

⠀⠀

$\bigstar\:{\underline{\boxed{\sf{\sqrt{\bigg(x_2 - x_1 \bigg)^2 + \bigg(y_2 - y_1 \bigg)^2}}}}}\qquad\qquad\bigg\lgroup\sf Distance\:formula\bigg\rgroup$

$\dashrightarrow\sf \sqrt{\bigg(x - 7\bigg)^2 + \bigg(y - 0\bigg)^2} = \sqrt{\bigg(x - 0\bigg)^2 + \bigg(y - 5\bigg)^2}\\\\\\ \longrightarrow\sf (x - 7)^2 + (y - 0)^2 = (x - 0)^2 + (y - 5)^2\\\\\\ \longrightarrow\sf\cancel{x^2} - 14x + 49 + \cancel{y^2} = \cancel{x^2} + \cancel{y^2} - 10y + 25\\\\\\ \longrightarrow\sf - 14x + 49 = - 10y + 25\\\\\\ \longrightarrow\sf - 14x + 10y = 25 - 49\\\\\\ \longrightarrow\sf - 2(7x - 5y) = - 24\\\\\\ \longrightarrow\sf 7x - 5y = \cancel{\dfrac{- 24}{-2}}\\\\\\ \longrightarrow{\underline{\boxed{\frak{\pink{7x - 5y = 12}}}}}\:\bigstar$

⠀⠀

$\therefore\:{\underline{\sf{Hence,\:the\:relation\:between\:x\:\&\:y\:is\: {\pmb{7x - 5y = 12}}.}}}$

Answer 2

Answer:

Given :-

a(7,0)

b(0,5)

To Find :-

Relation between x and y

Solution :-

We know that

$\sf \: D = \sqrt{ \bigg(x_{2} - x _{1} \bigg)^2 + \bigg(y _{2} - y _{1} \bigg)^2 }$

$\sf D = \sqrt{\bigg(x - 7\bigg)^2 + \bigg(y - 0\bigg)^2} = \sqrt{\bigg(x - 0\bigg)^2+\bigg(y - 5\bigg)^2}$

$\sf D = \bigg(x^2 - 14x + 49 + y^2\bigg) = \bigg(x^2 + y^2 - 10y + 25\bigg)$

$\sf \: D = {x}^{2} - 14x + 49 + y^2 = {x}^{2} + {y}^{2} - 10y + 25$

$\sf \: D = - 14x + 49 = - 10y + 25$

$\sf \:D = 49 - 25 = 14x + ( - 10y)$

$\sf \: D =24 = 14 x - 10y$

$\sf \: D = \dfrac{24}{2} = \dfrac{14x - 10y}{2}$

$\sf \: D =12 = 7x - 5y$