Question

Find the point $(0,y)$ that is equidistant from $(4,-9)$ and $(0,-2)$

• A. $(0,-\frac{93}{14})$
• B. $(\frac{93}{14},0)$
• C. $(0,-\frac{14}{93})$
• D. $(\frac{14}{93},0)$

Answer 1

The correct option is A $(0,-\frac{93}{14})$

Since the point $(0,y)$ is equidistant from the points $(4,-9)$ and $(0,-2)$,

therefore,

We find the distance between the points $(0,y)$ and $(4,-9)$ that is :

$D=\sqrt{{(0-4)}^2+{(y+9)}^2}$

$=\sqrt{(-4{)}^2+{(y+9)}^2}$

$=\sqrt{16+{(y+9)}^2}$

Now the distance between the points $(0,y)$ and $(0,-2)$ that is :

$D=\sqrt{{(0-0)}^2+{(y-(-2\left)\right)}^2}=\sqrt{{(y+2)}^2}$

Now equate the distances as follows:

$\sqrt{16+{(y+9)}^2}=\sqrt{{(y+2)}^2}$

Squaring both sides we get,

$16+{(y+9)}^2={(y+2)}^2$

$16+y^2+81+18y=y^2+4+4y$

$97+18y=4+4y$

$14y=-93$

$y=-\frac{93}{14}$

Hence, the point $(0,-\frac{93}{14})$ is equidistant from the points $(4,-9)$ and $(0,-2)$.