Question

Find the point on the y-axis which is equidistant from $A(3,-6)$ and $B(-2,5)$.

• A. $(\frac{9}{11},0)$
• B. $(0,\frac{8}{11})$
• C. $(0,\frac{-8}{11})$
• D. $(0,\frac{9}{11})$

Answer 1

The correct option is C $(0,\frac{-8}{11})$

Let the point be $(x,y)$.

Distance formula: $d^2=(x_1-x_2{)}^2+(y_1-y_2{)}^2$

Now equate the distances of the points $A(3,-6)$ and $B(-2,5)$ from the point $(x,y)$.

$(x-3{)}^2+(y+6{)}^2=(x+2{)}^2+(y-5{)}^2$

The $x-$coordinate of a point on $y-$axis is $0$.

$(0-3{)}^2+(y+6{)}^2=(0+2{)}^2+(y-5{)}^2$

$9+y^2+12y+36=4+y^2-10y+25$

$12y+10y=29-45$

$22y=-16$

$y=-\frac{8}{11}$

Hence, the point on the y-axis is $(0,-\frac{8}{11})$.