Show that the points A(1,5),B(-1,-1) and C (4,14) are collinear?

To show that the points A(1,5), B(-1,-1), and C(4,14) are collinear, we need to calculate the slope of the line passing through each pair of points and show that they are equal.

The slope of a line passing through two points (x1, y1) and (x2, y2) is given by the formula:

m = (y2 - y1) / (x2 - x1)

For points A(1,5) and B(-1,-1):

m_AB = (-1 - 5) / (-1 - 1) = -6 / -2 = 3

For points B(-1,-1) and C(4,14):

m_BC = (14 - (-1)) / (4 - (-1)) = 15 / 5 = 3

Since the slopes of the lines passing through AB and BC are equal, the points A, B, and C are collinear.

The points A(1,5),B(-1,-1) and C (4,14) are collinear if the area of the triangle formed by them is zero.
Area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by,
$$\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2 & 1 \\\ x_3 & y_3 & 1 \\\ \end{vmatrix}$$
Here, the given points are A(1,5),B(-1,-1) and C (4,14),
So, the area of the triangle ABC is
$$\frac{1}{2}\begin{vmatrix} 1 & 5 & 1 \\\ -1 & -1 & 1 \\\ 4 & 14 & 1 \\\ \end{vmatrix}$$
$$=\frac{1}{2}[1(-1-14)+5(-1-4)+1(5--4)]$$
$$=\frac{1}{2}(-15-25+9)$$
$$=\frac{1}{2}(-31)$$
$$=-\frac{31}{2}$$
Since the area of the triangle ABC is not zero, therefore, the points A(1,5),B(-1,-1) and C (4,14) are not collinear.