# 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

Verified

362.7k+ views

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.

Last updated date: 23rd Sep 2023

â€¢

Total views: 362.7k

â€¢

Views today: 4.62k

Recently Updated Pages

What do you mean by public facilities

Slogan on Noise Pollution

Paragraph on Friendship

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

What is the Full Form of ILO, UNICEF and UNESCO

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What is the past tense of read class 10 english CBSE