'How to find the most left point in a Triangle?

I have this problem: I have this class named Point:

class Point {
private double x;
private double y;

public Point(final double x, final double y) {
    this.x = x;
    this.y = y;
}

public double getX() {
    return x;
}

public double getY() {
    return y;
}

}

I have a class Triangle:

class Triangle {
private Point a;
private Point b;
private Point c;

public Triangle(Point a, Point b , Point c){
    this.a = a;
    this.b = b;
    this.c = c;
}

@Override
public double area() {
    double area = (a.getX() * (b.getY()- c.getY()) +
            b.getX() * (c.getY() - a.getY()) + c.getX() * (a.getY() - b.getY())) / 2.0;
    return Math.abs(area);
}

@Override
public String pointsToString() {
    return "("+a.getX()+","+a.getY()+")"+"("+b.getX()+","+b.getY()+")"+"("+c.getX()+","+c.getY()+")";
}

public String toString() {
    return this.getClass().getSimpleName() + "[" + pointsToString() + "]";
}

@Override
public Point leftmostPoint() {
    return null;
}

}

The question is how to find the leftmost point in the Triangle ?

I should realize the method leftmostPoint(). The method leftmostPoint() should return leftmost point of the Triangle: the one having the least X coordinate and if there are many leftmost points, return any of them.

javaoop


Solution 1:[1]

Think about how triangles work. The leftmost point (or one of them, if there are more than one) must be one of the 3 corners. So, it's a, b, or c.

Just compare a to b, and then the winner of that to c, and return the winner of that.

Solution 2:[2]

the least X coordinate of a Triangle will work

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 rzwitserloot
Solution 2 Eugene B

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