Question

Find the values of p for which the given points are collinear:

(i) A(−1, 3) B(2, p) and C(5, −1)
(ii) A(3, 2) B(4, p) and C(5, 3)
(iii) A(−3, 9) B(2, p) and C(4, −5)

Answer 1

(i) A(−1, 3), B(2, p) and C(5, −1) are the given points. Then:
(x₁ = −1, y₁ = 3), (x₂ = 2, y₂ = p) and (x₃ = 5, y₃ = −1)
It is given that points A, B and C are collinear.
$$\begin{gathered} \Rightarrow x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)=0 \\ \Rightarrow \left(-1\right)\left(p-\left(-1\right)\right)+2\left(-1-3\right)+5\left(3-p\right)=0 \\ \Rightarrow \left(-1\right)\left(p+1\right)+2\left(-4\right)+5\left(3-p\right)=0 \\ \Rightarrow -p-1-8+15-5p=0 \\ \Rightarrow 6-6p=0 \\ \Rightarrow 6p=6 \\ \Rightarrow p=\frac{6}{6}=1 \end{gathered}$$
Therefore, the value of p is 1.

(ii) A(3, 2) B(4, p) and C(5, 3) are the given points. Then:
(x₁ = 3, y₁ = 2), (x₂ = 4, y₂ = p) and (x₃ = 5, y₃ = 3)
It is given that points A, B and C are collinear.
$$\begin{gathered} \Rightarrow x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)=0 \\ \Rightarrow 3\left(p-3\right)+4\left(3-2\right)+5\left(2-p\right)=0 \\ \Rightarrow 3\left(p-3\right)+4\left(1\right)+5\left(2-p\right)=0 \\ \Rightarrow 3p-9+4+10-5p=0 \\ \Rightarrow 5-2p=0 \\ \Rightarrow 2p=5 \\ \Rightarrow p=\frac{5}{2} \end{gathered}$$
Therefore, the value of p is $\frac{5}{2}$.

(iii) A(−3, 9) B(2, p) and C(4, −5) are the given points. Then:
(x₁ = −3, y₁ = 9), (x₂ = 2, y₂ = p) and (x₃ = 4, y₃ = −5)
It is given that points A, B and C are collinear.
$$\begin{gathered} \Rightarrow x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)=0 \\ \Rightarrow \left(-3\right)\left(p-\left(-5\right)\right)+2\left(-5-9\right)+4\left(9-p\right)=0 \\ \Rightarrow \left(-3\right)\left(p+5\right)+2\left(-14\right)+4\left(9-p\right)=0 \\ \Rightarrow -3p-15-28+36-4p=0 \\ \Rightarrow -7-7p=0 \\ \Rightarrow 7p=-7 \\ \Rightarrow p=\frac{-7}{7}=-1 \end{gathered}$$
Therefore, the value of p is −1.