# FIND THE EQUATION OF PERPENDICULAR BISECTOR

Find the Equation of Perpendicular Bisector :

Here we are going to see how to find the equation of perpendicular bisector.

## Find the Equation of Perpendicular Bisector - Example

Question 10 :

Find the equation of the perpendicular bisector of the straight line segment joining the points (3,4) and (-1,2)

Solution :

midpoint of the line segment joining the points (3,4) and (-1,2) x₁ = 3, y₁ = 4, x₂ = -1 , y₂ = 2

=  [(x + x)/2 , (y₁ + y₂)/2]

=  (3 + (-1))/2 , (4 + 2)/2

=  (2/2, 6/2)

=  (1 , 3)

slope of the line joining the points (3,4) and (-1,2)

m  =  (y₂ - y₁)/(x₂ - x₁)

=  (2 - 4)/(-1-3)

=  -2/(-4)

=  1/2

Slope of the required line  =  -1/(1/2)

=  -2

Equation of the required line :

(y - y₁)/(y₂ - y₁) =   (x - x₁)/(x₂ - x₁)

(y - 4)/(2 - 4) = (x - 3)/(-1 - 3)

(y - 4)/(-2) = (x - 3)/(-4)

-4(y - 4) = -2(x - 3)

- 4 y + 16 = - 2 x + 6

2 x -4 y + 16 - 6 = 0

2 x - 4 y + 10 = 0

dividing the whole equation by 2, we get

x - 2 y - 5 =0

## Finding the Intersection of Two Lines and Parallel to the Line Passing Through the Points

Question 11 :

Find the equation of the straight line passing through the point of intersection of the lines 2x+y-3=0 and 5x  + y - 6 = 0 and parallel to the line joining the points (1,2) and (2,1)

Solution :

To find the point of intersection of any two lines we need to solve them

2x + y - 3 = 0 ---- (1)

5x  + y - 6 = 0 ---- (2)

(1) - (2)

2x + y - 3 = 0

5x  + y - 6 = 0

(-)  (-)  (+)

-----------------

- 3x + 3 = 0

- 3 x = -3

x = 1

Substitute x = 1 in the first equation

2(1) + y - 3 = 0

-1 + y = 0

y = 1

the point of intersection is (1,1)

The required line is parallel to the line joining the points (1, 2) and (2, 1). So their slopes will be equal.

m  =  (y₂ - y₁)/(x₂ - x₁)

=  (1-2)/(2-1)

=  -1/1  = -1

Slope of the required line is -1 and a point on the line is (1, 1)

Equation of the Line :

(y - y₁) = m (x - x₁)

(y - 1) = -1 (x - 1)

y - 1 = - x + 1

x + y - 1 - 1 = 0

x + y - 2 = 0 After having gone through the stuff given above, we hope that the students would have understood, how to find the equation of perpendicular bisector.

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

You can also visit our following web pages on different stuff in math.

WORD PROBLEMS

Word problems on simple equations

Word problems on linear equations

Algebra word problems

Word problems on trains

Area and perimeter word problems

Word problems on direct variation and inverse variation

Word problems on unit price

Word problems on unit rate

Word problems on comparing rates

Converting customary units word problems

Converting metric units word problems

Word problems on simple interest

Word problems on compound interest

Word problems on types of angles

Complementary and supplementary angles word problems

Double facts word problems

Trigonometry word problems

Percentage word problems

Profit and loss word problems

Markup and markdown word problems

Decimal word problems

Word problems on fractions

Word problems on mixed fractrions

One step equation word problems

Linear inequalities word problems

Ratio and proportion word problems

Time and work word problems

Word problems on sets and venn diagrams

Word problems on ages

Pythagorean theorem word problems

Percent of a number word problems

Word problems on constant speed

Word problems on average speed

Word problems on sum of the angles of a triangle is 180 degree

OTHER TOPICS

Profit and loss shortcuts

Percentage shortcuts

Times table shortcuts

Time, speed and distance shortcuts

Ratio and proportion shortcuts

Domain and range of rational functions

Domain and range of rational functions with holes

Graphing rational functions

Graphing rational functions with holes

Converting repeating decimals in to fractions

Decimal representation of rational numbers

Finding square root using long division

L.C.M method to solve time and work problems

Translating the word problems in to algebraic expressions

Remainder when 2 power 256 is divided by 17

Remainder when 17 power 23 is divided by 16

Sum of all three digit numbers divisible by 6

Sum of all three digit numbers divisible by 7

Sum of all three digit numbers divisible by 8

Sum of all three digit numbers formed using 1, 3, 4

Sum of all three four digit numbers formed with non zero digits

Sum of all three four digit numbers formed using 0, 1, 2, 3

Sum of all three four digit numbers formed using 1, 2, 5, 6 