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Coordinate Geometry
EXERCISE 7.1
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

Step-by-Step Solution

Key Concept: Three points are collinear if and only if the slopes of the line segments joining any two pairs of points are equal (or equivalently, the area of the triangle formed by them is zero).
1. Write the coordinates\
\[A(1,5),\; B(2,3),\; C(-2,-11)\]
2. Find the slope of AB\
\[m_{AB}=\frac{y_2-y_1}{x_2-x_1}=\frac{3-5}{2-1}=\frac{-2}{1}=-2\]
3. Find the slope of AC\
\[m_{AC}=\frac{y_3-y_1}{x_3-x_1}=\frac{-11-5}{-2-1}=\frac{-16}{-3}=\frac{16}{3}\]
4. Compare the slopes\
Since \(m_{AB} = -2\) and \(m_{AC}=\frac{16}{3}\) are not equal, the slopes are different.
5. Conclusion\
Because the slopes are not equal, the three points do not lie on a single straight line. Hence the points are not collinear.

*Alternative method (using determinant)*:\
The points are collinear if \[\begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix}=0.\]
Substituting the coordinates gives \[\begin{vmatrix} 1 & 5 & 1\\ 2 & 3 & 1\\ -2 & -11 & 1\end{vmatrix}=1\cdot(3\cdot1-1\cdot(-11)) -5\cdot(2\cdot1-1\cdot(-2)) +1\cdot(2\cdot(-11)-3\cdot(-2))\]
\[=1(3+11)-5(2+2)+1(-22+6) =14-20-16 = -22
eq 0.\]
Since the determinant is non‑zero, the points are not collinear.

Correct Answer: The points (1,5), (2,3) and (–2,–11) are not collinear.
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