Question

Find the values of (x,y) if the distance to the point X, Y from (-3, 0) as well as from (3,0) are 4

Answer 1

Answer:

(0,√7) and (0, -√7)

Step-by-step explanation:

Given that the point P(x, y) is at the same distance from both the points

A (-3, 0) and B (3,0)

Hence, the point P(x,y) should be on the perpendicular bisector of the line joining A and B.

(Note, for any point P on the perpendicular bisector PA = PB, geometrically also we can verify).

Given points are on the x-axis.

Midpoint of AB is Origin.

Hence the perpendicular bisector of AB is y-axis.

Hence P lies on the Y-axis , so x = 0

Now PA =PB =4 (given)

Using distance formula, we get

9 + y² = 16

y² = 7

=> y = ±√7

Answer 2

Answer:
coordinates of A (0,√7) and A (0,-√7)

Solution:

Distance Formula:

$\sqrt{ {(x_{2} - x_{1})}^{2} + ( {y_{2} - y_{1})}^{2} }$

let A (x,y) B(-3,0)

$AB = \sqrt{ {(x + 3)}^{2} + {y}^{2} } = 4 \\ \\ {x}^{2} + {y}^{2} + 9 + 6x = 16 \\ \\ {x}^{2} + {y}^{2} + 6x = 7 ..eq1$

A (x,y) C(3,0)

$AC = \sqrt{( {x - 3)}^{2} + {y}^{2} } = 4 \\ \\ {x}^{2} + {y}^{2} - 6x + 9 = 16 \\ \\ {x}^{2} + {y}^{2} - 6x = 7 ..eq2$

put the value of

${x}^{2} + {y}^{2} = 7 - 6x$
in eq 2

$7 - 6x - 6x + 7 = 0 \\ \\ - 12x = 0 \\ \\ x = 0$
put value of x in any equation to find the value of y

${y}^{2} = 7 \\ \\ y = ± \sqrt{7}$

So coordinates of A (0,√7) and A (0,-√7)