You are given an array of ‘POINTS’ in a 2-D cartesian plane, where “POINTS[i] = [Xi, Yi]”, in which ‘Xi’ denotes the x-coordinate of the ‘i’th point, and ‘Yi’ denotes the y-coordinate of the ‘i’th point.
Let’s define the term ‘Boomerang’ as the set of three distinct points that are not in a straight line. Your task is to find out whether these points are a “boomerang” or not.
Note :1. The size of the array ‘POINTS’ is always ‘3’.
2. For better understanding, you may look at the sample test cases.
The first line contains an integer ‘T’, which denotes the number of test cases to be run. Then, the T test cases follow.
The next ‘3’ line of each test case contains two positive integers, ‘Xi’, and ‘Yi’, denoting the x and y coordinates of the ith point.
Output Format :
For each test case, print ‘true’ if the given points are a “boomerang”. Otherwise, print ‘false’.
Output for each test case will be printed in a separate line.
Note :
You do not need to print anything. It has already been taken care of. Just implement the given function.
1 <= ‘T’ <= 10^5
POINTS.length == 3
POINTS[i].length == 2
-100 <= ‘Xi’, ‘Yi’ <= 100
Where ‘POINTS’ is an array of points, ‘Xi’ and ‘Yi’ are the i-th point’s x and y coordinates.
Time Limit: 1sec
2
0 0
1 1
2 2
0 0
1 1
2 1
false
true
Test Case 1 :
If you look at the graph, all three points are in a straight line.

Test Case 2 :
If you look at the graph, there is no straight line possible to pass through all three points.

2
4 0
4 5
2 1
2 2
2 4
2 6
true
false
Think of finding the area of a triangle.
O(1).
As we are using only a variable to store the area of the triangle. Hence, the time complexity will be O(1).
O(1).
As we are using constant extra space.