Question

For what value of y, are the points P(1, 4), Q(3, y) and R(−3, 16) collinear?

Answer 1

P(1, 4), Q(3, y) and R(−3, 16) are the given points. Then:
(x₁ = 1, y₁ = 4), (x₂ = 3, y₂ = y) and (x₃ = −3, y₃ = 16)
It is given that the points P, Q and R are collinear.
Therefore,
$$\begin{gathered} x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)=0 \\ \Rightarrow 1\left(y-16\right)+3\left(16-4\right)+\left(-3\right)\left(4-y\right)=0 \\ \Rightarrow 1\left(y-16\right)+3\left(12\right)-3\left(4-y\right)=0 \\ \Rightarrow y-16+36-12+3y=0 \\ \Rightarrow 8+4y=0 \\ \Rightarrow 4y=-8 \\ \Rightarrow y=-\frac{8}{4}=-2 \end{gathered}$$
When y = −2, the given points are collinear.