Question

Find the relationship between x and y so that (x, y) is equidistant from (-3, 4) and (0, 3).​

Answer 1

Answer:-

Given:

Distance between ( - 3 , 4) and (x , y) = Distance between (0 , 3) and (x , y).

We know that,

Distance between two points with coordinates (x₁ , y₁) and (x₂ , y₂) is :

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

$\implies \sf \sqrt{ {(x - ( - 3))}^{2} + {(y - 4)}^{2} } = \sqrt{( {x - 0)}^{2} + {(y - 3)}^{2} }$

On squaring both sides we get,

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

• (a + b)² = a² + b² + 2ab

• (a - b)² = a² + b² - 2ab

$\: \implies \sf \: \not{ {x}^{2}} + \not{{3}^{2} } + 2 \times x \times 3 + \not{ {y}^{2} }+ {4}^{2} - 2 \times y \times 4= \not{ x ^{2} } + \not{ {y}^{2}} + \not{ {3}^{2} } - 2 \times 3 \times y \\ \\ \implies \sf \:6x + 16 - 8y = - 6y \\ \\ \implies \sf \:6x = - 6y + 8y - 16 \\ \\ \implies \sf \:x = \frac{2y - 16}{6} \\ \\ \implies \sf \:x = \frac{2(y - 8)}{6} \\ \\ \implies \boxed{ \sf \:x = \frac{y - 8}{3} }$

Answer 2

Given :

• Since the point (x,y) is equidistant from the points (−3,4) and (0,3)

To Find :

• Find the relationship between x and y

Solution :

We find the distance between the points (x,y) and (0,3) that is :

D = (x- 0)² + (y - 3)² = x² + (y - 3)

Now the distance between the points (x,y) and (- 3 ,4) that is :

D = [x - ( -3 ) ]2 + (y - 4)² = (x + 3)² + (y - 4)

Now equate the distances as follows :

x² + (y - 3)² = (x + 3)² + (y - 4)²

Squaring both sides :

x² + (y - 3)² = (x + 3)² + (y - 4)²

x² + y² + 9 - 6y = x² + 9 + 6x + y² + 16 - 8y

- 6y = 6x + 16 - 8y

6x + 16 - 8y + 6y

6x + 16 - 2y

3x + 8 - y

x = y - 8 / 3