Question

Find the equation of perpendicular bbisector of the line segment joining the points (1,1) & (2,3) .​

Answer 1

Answer:

Understandable to Students whose fundamentals are not Clear!

To find Equation of any line we need atleast two things:

1-Slope m

2-any one point(x1,y1) on that line

we can then apply formula:(y-y1)=m(x-x1)

we can obtain both of these

let's calculate slope first:

slope when two points are given is (y2-y1)/(x2-x1)

4-(-2)/(6–2)

3/2 this is the slope of given line

we need slope of line perpendicular to this line.

multiplication of slopes of two perpendicular lines is (-1)

so 3/2 ×m=-1

m=-2/3

2 Now let's find a point on the perpendicular line:

Because the perpendicular line divides the given line in two equal halves,

Perpendicular line passes through the mid point of given line.

Mid point of line, joining two points (x1,y1) (x2,y2) is

(x1+x2)/2 ,(y1+y2)/2

putting values (2,-2) (6,4)

(6+2)/2 , (4+-2)/2

8/2 , 2/2

(4,1) this is the mid point through which the second line passes

We can now use the formula y-y1=m(x-x1)

and put value -2/3 for m and (4,1) for (x1,y1)

y-1=-2/3(x-4) this is the required equation in point slope form.

we can shift 3 to LHS

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

3y-3=-2x+8

after rearrangement we get

2x+3y=11 this is the required equation in slope standard form

Thanks