Question

The value of $ p $ for which the points $(-5, 1)$, $(1, p)$, and $(4, -2)$ are collinear is:

• A. $ 1 $
• B. $ -2 $
• C. $ -1 $
• D. $ 3 $

Hint:

For three points to be collinear, the area of the triangle formed by them must be zero. Use the determinant condition to set up an equation and solve for the unknown variable. Always simplify step-by-step to avoid errors.

Answer 1

The Correct Option is C

Step 1: Understand Collinearity Condition. 
Three points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ are collinear if the area of the triangle formed by these points is zero. This can be expressed using the determinant condition: $$ \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0. $$ Substituting the given points $(-5, 1)$, $(1, p)$, and $(4, -2)$ into the determinant: $$ \begin{vmatrix} -5 & 1 & 1 \\ 1 & p & 1 \\ 4 & -2 & 1 \end{vmatrix} = 0. $$ Step 2: Expand the Determinant. 
Expanding along the first row: $$ \begin{vmatrix} -5 & 1 & 1 \\ 1 & p & 1 \\ 4 & -2 & 1 \end{vmatrix} = -5 \begin{vmatrix} p & 1 \\ -2 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 4 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & p \\ 4 & -2 \end{vmatrix}. $$ Calculate each minor determinant:
1. $\begin{vmatrix} p & 1 \\ -2 & 1 \end{vmatrix} = p(1) - (-2)(1) = p + 2$,
2. $\begin{vmatrix} 1 & 1 \\ 4 & 1 \end{vmatrix} = 1(1) - 4(1) = 1 - 4 = -3$,
3. $\begin{vmatrix} 1 & p \\ 4 & -2 \end{vmatrix} = 1(-2) - 4(p) = -2 - 4p$.
Substitute back into the expanded determinant: $$ -5(p + 2) - 1(-3) + 1(-2 - 4p) = 0. $$ Simplify: $$ -5p - 10 + 3 - 2 - 4p = 0, $$ $$ -5p - 4p - 10 + 3 - 2 = 0, $$ $$ -9p - 9 = 0. $$ Solve for $ p $: $$ -9p = 9 \implies p = -1. $$
Step 3: Analyze the Options. 
Option (1): $ 1 $ — Incorrect, as $ p = -1 $. 
Option (2): $ -2 $ — Incorrect, as $ p = -1 $. 
Option (3): $ -1 $ — Correct, as this matches the calculated value. 
Option (4): $ 3 $ — Incorrect, as $ p = -1 $.
Step 4: Final Answer. (3) \(\mathbf{-1}\)