# 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

Last updated date: 19th Mar 2023

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Answer

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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.

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.

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