# Show that the points A(-1,2), B(5,2) & C(2,5) are the vertices of an isosceles triangle. Find the area of triangle ABC.

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By definition, an isosceles triangle is one that has two congruent (equal) angles. To prove two angles equal on the triangle ABC, let us consider the definition of tangent:

tanx = opposite/adjacent = h/a

Noting that the y values for points A and B are equal, the height is the distance from point C to the line y = 2:

h = 5 - 2 = 3

The distance AB = 5 - -1 = 6

Let D be the point through C perpendicular to AB; i.e. the line x = 2.

AD = 2 - -1 = 3 ; DB = 5 - 2 = 3

(note that a triangle with perpendicular bisector is isosceles. But proof by definition follows)

The tangent of angle CAB = h/AD = 3/3 = 1

The tangent of angle CBA = h/DB = 3/3 = 1

Therefore the angles are congruent, and the triangle is isosceles.

The definition of the Area of a triangle is half base times height. We know both from above:

h = 3, AB = 6

0.5*h*AB = 9

Let us calulate the length of the triangle sides:

AB = sqrt[(2-2)^2 + (5+1)^2]= sqrt(36) = 6

BC = sqrt[(5-2)^2 + (2-5)^2]= sqrt18 = 3sqrt2

AC = sqrt[(5-2)^2 + (2+1)^2 = sqrt18 = 3sqrt2

We notice that BC = AC = 3sqrt2

Then ABC is an isoscale triangle.

Now The area:

We know that the area a= (1/2)base * height

height is CD where D in the midpoint of AB

==> CD^2 = BC^2 - (AB/2)^2

==> CD = sqrt(18 - 9) = sqrt9 = 3

==> a = (1/2)AB * CD

= (1/2)*6 * 3

= 3*3

==> area (a) = 9

A )(-1,2), B(5,2) and C(2,5)

Therfore AB^2 = (5- -1)^2+(2-2)^2 = (5+2)^2 = 49. Or AB = sqrt(49) = 7.

BC^2 = = (2-5)^2+(5-2)^2 = 9+9=18. BC = sqrt(18) = 3sqrt2.

AC^2 = (2- -1)^2+(5-2)^2= 9+9= 18. AC = sqrt18 = 3sqrt2.

Therefore BC =AC = 3sqrt2. Triangle ABC is an isosceles .

Therefore square on altitude from AB to C = AD^2 = AC^2 - { (1/2) AB}^2 = 18 - 49/4 = 18- 12.25 = 5.75.

Therefore the area of the triangle = (1/2) *BC*AD = (1/2)*7* sqrt5.75 = 8.39271 nearly.