Question

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

Answer 1

Answer:

3x - y = 5

Step-by-step explanation:

Let A(3,6) and B(-3,4)

Now, Let P(x,y) be equidistant to A and B

Thus,

PA = PB

PA² = PB²

Use the distance formula - $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Since PA and PB are squared, the square root cancels

(x-3)² + (y-6)² = (x+3)²+(y-4)²

(x²+ 9 - 6x) + (y² + 36 - 12y) = (x² + 9 + 6x) + (y² + 16 - 8y)

45 - 6x - 12y = 25 + 6x - 8y

20 = 12x - 4y

3x - y = 5

Answer 2

Answer:

Question

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

Answer

Given :-

• Two points A(3, 6) and B(- 3, 4)

To find :-

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

Formula used :-

• Distance formula :- This formula is used to find the distance between two points and formula to find distance between two ponts A and B is given by

$\bf \:AB = \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} }$

Solution:-

Since P(x, y) is equidistant from A(3, 6) and B(- 3, 4).

$\bf\implies \:PA = PB$

Squaring both sides, we get

$\bf\implies \: {PA}^{2} = {PB}^{2}$

$\bf\implies \: {(x - 3)}^{2} + {(y - 6)}^{2} = {(x + 3)}^{2} + {(y - 4)}^{2}$

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

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

$\bf\implies \:12x + 4y = 20$

On divides both sides by 4

$\bf\implies \:3x + y = 5$