Step 1: Understanding the Concept:
Three points are collinear if they lie on the same straight line. This can be verified by showing that the slope of the line segment connecting any two points is the same as the slope of the line segment connecting any other two points.
Step 2: Key Formula or Approach:
The slope \(m\) of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
We will calculate the slope of AB and the slope of BC. If they are equal, the points are collinear.
Step 3: Detailed Explanation:
The given points are A(1, -3), B(2, -5), and C(-4, 7).
1. Find the slope of line segment AB:
Here, \((x_1, y_1) = (1, -3)\) and \((x_2, y_2) = (2, -5)\).
\[ m_{AB} = \frac{-5 - (-3)}{2 - 1} = \frac{-5 + 3}{1} = \frac{-2}{1} = -2 \]
2. Find the slope of line segment BC:
Here, \((x_1, y_1) = (2, -5)\) and \((x_2, y_2) = (-4, 7)\).
\[ m_{BC} = \frac{7 - (-5)}{-4 - 2} = \frac{7 + 5}{-6} = \frac{12}{-6} = -2 \]
Since the slope of AB is equal to the slope of BC (\(m_{AB} = m_{BC} = -2\)), and they share a common point B, the points A, B, and C lie on the same straight line.
Step 4: Final Answer:
Yes, the points A, B, and C are collinear.
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\). 
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\)) 
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.