Let us apply the distance formula to find the lengths PQ,QR, and PR, where P(3,2),Q(-2,-3) and R(2,3) are the given points.We have
PQ=√((--)2 + (--)2)
=√((-)2 + (-)2)
=√
=.07units(approx)
QR=√((-(-))2 + (-(-))2)
=√(2 + 2)
=√
=.21units(approx)
PR=√((-)2 + (-)2)
=√((-)2 + 2)
=√
=.41units(approx)
Since the sum of any two of these lengths is greater than the third length, the points P,Q and R from a triangle and all the sides of triangle are unequal.