The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find :

(i) the gradient of PQ;

(ii) the equation of PQ;

(iii) the co-ordinates of the point where PQ intersects the x-axis.

(i) By formula,

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

Substituting values we get,

$\text{Slope of PQ } = \dfrac{5 - 6}{-3 - 2} \\[1em] = \dfrac{-1}{-5} \\[1em] = \dfrac{1}{5}.$

Hence, gradient of slope PQ = $\dfrac{1}{5}$.

(ii) By point slope form,

Equation : y - y₁ = m(x - x₁)

⇒ y - 6 = $\dfrac{1}{5}$(x - 2)

⇒ 5(y - 6) = x - 2

⇒ 5y - 30 = x - 2

⇒ 5y = x - 2 + 30

⇒ 5y = x + 28

Hence, equation of PQ is 5y = x + 28.

(iii) The point where PQ intersects x-axis, there y co-ordinate = 0.

Substituting y = 0 in equation of PQ we get,

⇒ 5 × 0 = x + 28

⇒ 0 = x + 28

⇒ x = -28.

Point = (x, y) = (-28, 0).

Hence, co-ordinates of the point where PQ intersects the x-axis = (-28, 0).