Determining the Value of m for Collinear Points: A Comprehensive Guide

When working with points in coordinate geometry, one fundamental concept is determining whether given points are collinear. Collinear points lie on the same line, and this can be checked using various methods, one of which involves calculating the area of a triangle formed by these points. If the area is zero, the points are collinear. This article provides a detailed step-by-step process to find the value of m that makes the points (3, 5), (m, 6), and (left(frac{1}{2}, frac{15}{2}right)) collinear.

Conceptual Understanding: Area of a Triangle and Collinearity

The area A of a triangle formed by points ((x_1, y_1)), ((x_2, y_2)), and ((x_3, y_3)) can be calculated using the formula:

A (frac{1}{2} |x_1y_2 - y_3 - x_2y_3 y_1 - x_3y_1 y_2|)

Applying the Concept to the Given Points

Given points are (3, 5), (m, 6), and (left(frac{1}{2}, frac{15}{2}right)). We substitute these into the area formula:

A (frac{1}{2} |3 cdot 6 - frac{15}{2} - m cdot frac{15}{2} 5 - frac{1}{2} cdot 5 6|)

Let's break down each term:

First term: (3 cdot 6 - frac{15}{2} 18 - frac{15}{2} frac{36}{2} - frac{15}{2} frac{21}{2}) Second term: (-m cdot frac{15}{2} 5 -frac{15m}{2} 5) Third term: (-frac{1}{2} cdot 5 6 -frac{5}{2} 6 -frac{5}{2} frac{12}{2} frac{7}{2})

Combining these terms, we have:

A (frac{1}{2} | frac{21}{2} - frac{15m}{2} 5 - frac{5}{2} 6 |)

Simplifying inside the absolute value:

A (frac{1}{2} | frac{21}{2} - frac{15m}{2} frac{10}{2} - frac{5}{2} frac{12}{2} | frac{1}{2} | frac{38 - 15m}{2} | frac{1}{4} | 38 - 15m |)

For the points to be collinear, the area must be zero:

(frac{1}{4} | 38 - 15m | 0)

This implies:

(38 - 15m 0)

Solving for m gives:

(15m 38)

(m frac{38}{15})

However, reviewing the earlier calculation and more directly solving using slope method, we get:

Slope Method for Collinearity

The slope between the points (3, 5) and (left(frac{1}{2}, frac{15}{2}right)) is calculated as:

(text{slope} frac{frac{15}{2} - 5}{frac{1}{2} - 3} frac{frac{15}{2} - frac{10}{2}}{frac{1}{2} - frac{6}{2}} frac{frac{5}{2}}{-frac{5}{2}} -1)

The equation of the line using the point-slope form with point (3, 5) is:

(y - 5 -1(x - 3))

(y - 5 -x 3)

(y -x 8)

Substituting the point (m, 6) into this equation, we find:

6 -m 8

(-m -2)

(m 2)

Thus, the value of m that makes the points collinear is boxed{2}.