# 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) $(0,-13)$(C) $(0,12)$(D) $(13,0)$

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Given:

The line segment joining the points $A(1,5)$ and $B(4,6)$.

To do:

We have to find the point at which the perpendicular bisector of the line segment joining the points $A(1,5)$ and B $(4,6)$ cuts the $y$-axis.

Solution:

We know that,

The perpendicular bisector of a line segment divides the line segment into two equal parts.

The perpendicular bisector of the line segment passes through the mid-point of the line segment.

Let the perpendicular bisector of $A B$ meets $\mathrm{y}$ axis at $\mathrm{P}(0, \mathrm{y})$

Therefore,

$\mathrm{AP}=\mathrm{BP}$

Squaring both sides, we get,

$AP^{2}=BP^{2}$

Using distance formula, we get,

$(\mathrm{x}_{1}-0)^{2}+(\mathrm{y}_{1}-\mathrm{y})^{2}=(\mathrm{x}_{2}-0)^{2}+(\mathrm{x}_{2}-\mathrm{y})^{2}$

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

$1+5^2-2(5)(y)+y^2=16+6^2-2(6)(y)+y^2$

$1+25-10y=16+36-12y$

$12y-10y=52-26$

$2y=26$

$y=13$

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

Updated on 10-Oct-2022 13:28:26