Question

Find the point on the y-axis which is equidistant from A (3, - 4) and B (- 5, 9).

• A. $(0,\frac{81}{28})$
• B. $(0,3)$
• C. $(0,\frac{81}{26})$
• D. $(0,\frac{105}{26})$

Answer 1

The correct option is C

$(0,\frac{81}{26})$

We have to find a point on the y-axis. Therefore, its x-coordinate will be 0.
Let the point on y-axis be P (0,y)

Distance PA = Distance between $(0,y)$ and $(3,-4)=\sqrt{(0-3{)}^2+(y-(-4){)}^2}=\sqrt{(-3{)}^2+(y+4{)}^2}$

Distance PB = Distance between $(0,y)$ and $(-5,9)=\sqrt{(0-(-5){)}^2+(y-9{)}^2}=\sqrt{5^2+(y-9{)}^2}$

By the given condition, PA = PB, thus
$\sqrt{(-3{)}^2+(y+4{)}^2}=\sqrt{5^2+(y-9{)}^2}$

Squaring on both sides of the equation

$\therefore$ $(y+4{)}^2+9=(y-9{)}^2+25$

$y^2+16+8y+9=y^2-18y+81+25$

$26y=106-25$

$26y=81$

$y=\frac{81}{26}$

$\therefore$ The required point is $(0,\frac{81}{26})$.