Question

15. Find a relation between x and y such that the point (x, y) is equidistant from the points(-2, 8) and (-3,-5)​

Answer 1

Step-by-step explanation:

   Using distance formula:

⇒ √(-2 - x)² + (8 - y)² = √(-3 - x)² + (-5 - y)²

⇒ (-2 - x)² + (8 - y)² = (-3 - x)² + (-5 - y)²

⇒ (2 + x)² - (3 + x)² = (5 + y)² - (8 - y)²

⇒ (2 + x + 3 + x)(2 + x - 3 - x) = (5 + y + 8 - y)(5 + y - 8 + y)

⇒ (5 + 2x)(-1) = (13)(-3 + 2y)

⇒ - 5 - 2x = - 39 + 26y

⇒ 2x + 26y = 34

⇒ x + 13y = 17

Answer 2

Given :-

• The point (x, y) is equidistant from the points (- 2, 8) and (- 3, - 5)

To Find :-

• A relation between ‘x’ and ‘y’

Formula :-

We know that,

$\boxed{\sf\sqrt{(x_1 - x_2)^{2} +(y_1 - y_2)^{2}}} \: \: (\bf Distance \: Formula)$

Solution :-

$:\implies\sf \sqrt{ { (- 2 - x)}^{2} + {(8 - y)}^{2} } = \sqrt{ {( - 3 - x)}^{2} + ( - 5 - y)^{2} }$

$:\implies\sf {( - 2 - x)}^{2} + {(8 - y)}^{2} = {( - 3 - x)}^{2} + {( - 5 - y)}^{2}$

$:\implies\sf {(2 + x)}^{2} - {(3 + x)}^{2} = {(5 + y)}^{2} - {(8 - y)}^{2}$

$:\implies\sf(2 + x + 3 + x)(2 + x - 3 - x) = (5 + y + 8 - y)(5 + y - 8 + y)$

$:\implies\sf(5 + 2x)( - 1) = (13)( - 3 + 2y)$

$:\implies\sf - 5 - 2x = - 39 + 26y$

$:\implies\sf - 2x - 26y = - 39 + 5$

$:\implies\sf - 2x - 26y = - 34$

$:\implies\sf - 2\:(x + 13y) = - 34$

$:\implies\sf x + 13y = \frac{\cancel{- 34}}{\cancel{- 2}}$

$:\implies\sf x + 13y = 17$