Question

Find the value of X,Y if the distances of the point (X,Y) from (-3,0) as well as from (3,0) are 4.

Answer 1

Answer:-

Given:

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

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

We know that ,

$\sf{Distance \: between\: two\: points \:}= \sf{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$

$→\sf{\sqrt{(-3-x)^2 + (0 - y)^2}=4}\\\\→\sf{(-3-x)^2+y^2=16}\\\\→\sf{9 + x^2-6x+y^2=16}\\\\→\sf{y^2=16 -9- x^2 +6x}\\\\→\sf{y^2=7 - x^2 + 6x\:-\: equation\:(1)}$

Similarly,

$\sf{ \sqrt{( { 3 - x })^{2} + {(0 - y)}^{2} } = 4} \\ \\ → \sf{ {(3 - x)}^{2} + {y}^{2} = 16}$

Substitute "(y²)" value here.

$→ \sf{9 + {x}^{2} - 18x + 7 - {x}^{2} + 6x = 16 }\\ \\ →\sf{ 16 - 12x = 16} \\ \\ → \sf{ - 12x = 0} \\ \\ →\sf{x = 0}$

Substitute "x" value in equation (1)

$→\sf{ {y}^{2} = 7 - {(0)}^{2} + 6(0)} \\ \\ →\sf{y = 7}$

Hence, the Values of x and y are (0 , 7).