Question

Find the equation of the perpendicular bisector of $AB$ where $A$ and $B$ are the points $\left( {3,6} \right)$and $\left( { - 3,4} \right)$ respectively. Also, find its point of intersection with
     i) $x$-axis
     ii) $y$-axis

Answer 1

Hint: Since, it is given that it is a perpendicular bisector therefore, it will divide the line into two equal parts. Take a point P such that the distance PA=PB and then solve this equation using a line formula.

It is given that $A = \left( {3,6} \right)$ and$B = \left( { - 3,4} \right)$.
Now, we know that the equation of the perpendicular bisector of $AB$ will be the locus of the points which will be equidistant from $A$ and $B$.

Let us assume a point P(x,y) on the perpendicular bisector.
Therefore,
$PA = PB$
To simplify the formula of line formula we are going to square both sides,
${\left( {PA} \right)^2} = {\left( {PB} \right)^2}$

Note: Make sure you take square of the terms only once.

On applying the simplified line formula, we get,
${\left( {x - 3} \right)^2} + {\left( {y - 6} \right)^2} = {\left( {x + 3} \right)^2} + {\left( {y - 4} \right)^2}$

On opening the brackets, we get,
${x^2} + {3^2} - 2\left( 3 \right)\left( x \right) + {y^2} + {6^2} - 2\left( 6 \right)\left( y \right) = {x^2} + {3^2} + 2\left( 3 \right)\left( x \right) + {y^2} + {4^2} - 2\left( y \right)\left( 4 \right)$
On cancelling the common terms,
$9 - 6x + 36 - 12y = 9 + 6x + 16 - 8y$
Again cancelling the common terms on the above steps, we get,
$12x + 4y - 20 = 0$

Taking 4 out common from all the terms,
$4\left( {3x + y - 5} \right) = 0$

Therefore, we can say that,
$3x + y - 5 = 0$ is the required equation.
Therefore, the equation of the perpendicular bisector of $AB$ where $A$ and $B$ are the points $\left( {3,6} \right)$and $\left( { - 3,4} \right)$ respectively.