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

The two given points are P(5, 1), Q(1, -1).

Slope of the line = $\dfrac{y$2 - y1}{x2 - x1}

$= \dfrac{-1 - 1}{1 - 5} \\[1em] = \dfrac{-2}{-4} \\[1em] = \dfrac{1}{2}.$

So, the equation of PQ is

⇒ y - y₁ = m(x - x₁)
⇒ y - 1 = $\dfrac{1}{2}(x - 5)$
⇒ 2(y - 1) = x - 5
⇒ 2y - 2 = x - 5
⇒ x - 2y - 5 + 2 = 0
⇒ x - 2y - 3 = 0.

Now if point R(11, 4) is collinear to points P and Q then it will satisfy the equation x - 2y - 3 = 0,

Putting values in L.H.S of the equation

$= 11 - 2(4) - 3 \\[1em] = 11 - 8 - 3 \\[1em] = 0.$

The equation of the line PQ is x - 2y - 3 = 0. Since, L.H.S. = 0 = R.H.S, thus R satisfies the equation. Hence, points P, Q and R are collinear.