Question

find the area of the triangle??

Answer 1

y-x=0......(1)
x+y=0..........(2)
x-k=0..............(3)
From (1) and (2) , we get;
x=0 and y=0
From (2) and (3) , we get;
x=k and y= (-k) and
From (1)and (3) we get;
x=k and y=k
Let A(k,k) , B(0,0) and C(k,-k) represent the vertices of the triangle ABC .
AB=sq.root of (k-0)^2 + (k-0)^2
=k*root2
AC=sq.root of (k-k)^2 + (k+k) ^2
AC =2k
and,
BC=sq.root of (k-0)^2 + (-k-0)^2
BC=2*root k
AB = BC i.e. the triangle is iso.triangle
Let BD is the perpendicula on sideAC

So,BD =sq.root of (AB^2 - BD^2)
i.e .BD = root (4k - k^2)
So , area of triangle ABC = 1/2*AC*BD
=1/2*2k*root(4k - k^2)
=k*root(4k - k^2) .