Find the area of the triangle ABC such that A(1,1) B(6,1) and C(1,9)

To find the are of the triangle A(1,1), B(6,1) and
C(1,9):

Solution:

We know by
distance formula that the distance between (x1,y1) and (x2,y2) = sqrt
{(x2-x1)^2+(y2-y1)^2}

A(1,1) , B(6,1): Therefore AB =
sqrt{(6-1)^2+(1-1)^2} = 5.

A(1,1) , C (1, 9) , therefore AC
=sqrt{ (1-1)^2+(9-1)^2} = 8.

Also ABC is right angle at A
as  AB is parallel to x axis , the equation of AB being y =
1.

Similarly AC has the equation , x = 1 . So AC is || to Y
axis.

Therefore AB and AC is Perpendicular to each
other.

So ABC is right angle at
A.

Theefore the area of ABC = (1/2)AB*AC = (1/2)(5*8) = 20
sq units.