A(1,1)

B(6,1)

C( 1,9)

To find the area we need to determine the lenght of all sides:

l AB l = sqrt(6-1)^2 + (1-1)^2

= sqrt(25) = 5

l AC l = sqrt(1-1)^2 + (9-1)^2

= sqrt(8^2 = 8

l BC l = sqrt( 6-1)^2 + (1-9)^2

= sqrt( 25+64) = sqrt 89

We notice that:

l AB l ^2 + l AC l ^2 = l BC l ^2

25 + 64 = 89

89 = 89

Then we conclude that ABC is a right angle triangle where BC is the hypotenuse.

Then the base and height are 5 and 8

Then the area is:

A = 1/2 * base * height

= 1/2 * 5 * 8

= 20

**Then the area of the triangle = 20 square units.**

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.