Question

Find the equation of the line passing through the points $P\left(5,1\right)$ and $Q\left(1,-1\right)$. Hence, show that the points $P,Q$ and $R\left(11,4\right)$ are collinear.

Answer 1

Step1: Calculation of equation of the line passing through $P(5,1),Q(1,-1)$.

The two-point form of the equation of a line passing through $\left(x_1,y_1\right)$and $\left(x_2,y_2\right)$ is given by the formula $\left(y-y_1\right)=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}(x-x_1)$.

Thus, the equation of a line passing through $P(5,1),Q(1,-1)$ is given by:

$$\begin{gathered} (y-1)=\frac{\left(-1-1\right)}{\left(1-5\right)}\left(x-5\right) \\ \Rightarrow (y-1)=\frac{\left(-2\right)}{\left(-4\right)}\left(x-5\right) \\ \Rightarrow (y-1)=\frac{1}{2}\left(x-5\right) \\ \Rightarrow 2y-2=x-5(multiplyingbothsidesby2) \\ \Rightarrow 0=x-5-(2y-2)(Subtracting2y-2frombothsides) \\ \Rightarrow 0=x-5-2y+2 \\ \therefore x-2y-3=0\left(rearranging\right) \end{gathered}$$

Step2: Proof that the points $P,Q$ and $R\left(11,4\right)$ are collinear.

Now, if point $R\left(11,4\right)$ is collinear to points $P$, $Q$ , then $R\left(11,4\right)$ should satisfy the equation of the line $PQ$.

The equation of line $PQ$ is $x-2y-3=0$

Thus,

$$\begin{gathered} 11-2\left(4\right)-3\stackrel{?}{=}0 \\ \Rightarrow 11-8-3\stackrel{?}{=}0 \\ \Rightarrow 0=0 \end{gathered}$$

Since, the point $R\left(11,4\right)$ satisfies the equation $x-2y-3=0$; Thus the points $P(5,1),Q(1,-1)$ and $R(11,4)$ are collinear.

Hence, the equation of line through the collinear points $P(5,1),Q(1,-1)$ and $R(11,4)$ is $x-2y-3=0$.