Question

exercise. show that the points \\(a(0,-2)\\), \\(b(5,1)\\) and \\(c(-5,-5)\\) are collinear

Explanation:

Identify the given points

We are given three points:
\(A(0, -2)\), \(B(5, 1)\), and \(C(-5, -5)\).

Calculate the slope of line segment AB

Using the Linear Equations concept, we find the slope \(m_{AB}\):

$$ m_{AB} = \frac{1 - (-2)}{5 - 0} = \frac{3}{5} $$

Calculate the slope of line segment BC

Using the Linear Equations concept, we find the slope \(m_{BC}\):

$$ m_{BC} = \frac{-5 - 1}{-5 - 5} = \frac{-6}{-10} = \frac{3}{5} $$

Compare the slopes to show collinearity

Since \(m_{AB} = m_{BC} = \frac{3}{5}\), the lines \(AB\) and \(BC\) are parallel.
Since they share the common point \(B\), the points \(A\), \(B\), and \(C\) must lie on the same straight line.
Thus, the points are collinear.

Answer:

To show that the points \(A(0, -2)\), \(B(5, 1)\), and \(C(-5, -5)\) are collinear, we calculate the slopes between the points:

• Slope of \(AB\):

$$m_{AB} = \frac{1 - (-2)}{5 - 0} = \frac{3}{5}$$

• Slope of \(BC\):

$$m_{BC} = \frac{-5 - 1}{-5 - 5} = \frac{-6}{-10} = \frac{3}{5}$$

Since \(m_{AB} = m_{BC}\) and the segments share the common point \(B\), the points \(A\), \(B\), and \(C\) lie on the same line and are therefore collinear.