Question

The perpendicular bisector of the line segment joining the points $A(1,5)$ and $B(4,6)$ cuts the $y$-axis at

• A. $(0,13)$
• B. $(0,14)$
• C. $(0,12)$
• D. $(13,0)$

Answer 1

The correct option is A $(0,13)$

The given points are $A(x_1,y_1)=(1,5)$ and $B(x_2,y_2)=(4,6)$.

The perpendicular bisector of $AB$ will pass through the midpoint of $AB$.

Now the perpendicular bisector of $AB$ will meet $Y$ axis at $P(0,y)$.

$\therefore AP=BP$

Applying distance formula, we get

$\Rightarrow A{P}^2=B{P}^2$

$\Rightarrow (x_1-0{)}^2+(y_1-y{)}^2=(x_2-0{)}^2+(x_2-y{)}^2$

$\Rightarrow (1-0{)}^2+(5-y{)}^2=(4-0{)}^2+(6-y{)}^2$

After solving, we get $y=13$

Therefore, the point is $(0,13)$.