Question

Find the point P(x, y) if its distance from (-3, 0) & (3, 0) is 4 units individually

• A. $(0,\sqrt{5})$
• B. $(0,-\sqrt{5})$
• C. $(0,-\sqrt{7})$
• D. $(1,0)$

Answer 1

The correct option is C $(0,-\sqrt{7})$

Distance
between two points $(x_1,y_1)$ and $(x_2,y_2)$ can be calculated using the formula $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
Given,
Distance
between the points $(x,y);(-3,0)=$ 4
between $(x,y);(3,0)=4$

Solving,we get
$=>12x=0$

$=>x=0$
Now,
Distance
between the points $(x,y);(-3,0)=4$

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

But $x=0$
$=>\sqrt{9+y^2}=4$
$=>9+y^2=16$

$y^2=7$

$=>y=\sqrt{7}$ or $-\sqrt{7}$