Vikas TU
Last Activity: 7 Years ago
Area of the triangle from all the three points in detreminant form is:
|h k 1|
|2 2 1| = 12
|-4 5 1|
On solving we get,
h(2-5) – k(2+4) + 1(10+8) = 12
-3h -6k = – 6
or
h + 2k = 2
x + 2y – 2 = 0 is the eqn. and the locus in terms of x and y respectively which is a straight line.