Question

23. The coordinates of ABC are A(1, 2), B(6, 7) and C(7,2). Find the equations of the perpendicular bisectors of (a)AB (b) BC.Hence find the coordinates of the point equidistant from A, B and C.​

Answer 1

Answer:

A-(1,2) , B-(6,7) and, C-(7,2)

To find the perpendicular bisectors we'll first find the midpoints of all the sides and the slopes of the lines perpendicular to the sides.

So, Midpoint of AC-(1+7/2,2+2/2) - (4,2)

Midpoint of BC - ( 13/2 , 9/2)

Midpoint of AB-(7/2,9/2)

Slopes of Sides AB - (7-2/6-1) = 5/5 = 1

of BC = (7-2/6-7) = -5

of CA =(2-2/7-1) = 0

So, slopes of lines perpendicular to these sides will be

-1, 1/5 , infinity

And, now hence we can find the equation of the three perpendicular bisectors

Perpendicular bisector of side AB :

(y-9/2) = -1(x-7/2)

Perpendicular bisector of side BC :

(y-9/2) = 1/5(x-13/2)

Perpendicular bisector of side AC:

(y-2)=1/0(x-4) => y=2

Hence, finding the intersection of any two of these perpendicular bisectors will give us a point which is also the circumcentre of this triangle ABC and hence the point that will actually be the point equidistant from all the three vertices and this distance will be equal to the circumadius of the traingle.

I'll leave to you to find the intersection of any of the two perpendicular bisectors.

After solving any two linear equations the answer of this point will be ( 6,2) as the circumcentre.