Question

Find the point on the y-axis which is equidistant from A (3, - 4) and B (- 5, 9). $\left[3\text{Marks}\right]$

Answer 1

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) $\left[0.5\text{Marks}\right]$

Distance PA = Distance between $P(0,y)$ and $A(3,-4)=\sqrt{(0-3{)}^2+(y-(-4){)}^2}=\sqrt{(-3{)}^2+(y+4{)}^2}$
$\left[0.5\text{Marks}\right]$

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

$\left[0.5\text{Marks}\right]$

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

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

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

$26y=106-25$

$26y=81$

$y=\frac{81}{26}$ ​​​​​​ $\left[1\text{Mark}\right]$

$\therefore$ The required point is $(0,\frac{81}{26})$. $\left[0.5\text{Marks}\right]$