A straight line passes through the points A(2, –4) and B(5, –2). Find :

(i) the slope of the line AB

(ii) the equation of the line AB

(iii) the value of k, if AB passes through the point P(k + 3, k – 4)

(i) Given points, A(2, –4) and B(5, –2)

$\Rightarrow \text{ Slope of AB } = \dfrac{y$2 - y1}{x2 - x1} \\[1em] = \dfrac{-2 - (-4)}{5 - 2} \\[1em] = \dfrac{2}{3}.

Hence, slope = $\dfrac{2}{3}$.

(ii) By point-slope form,

Equation of the line AB, y - y₁ = m(x - x₁)

⇒ y - (-4) = $\dfrac{2}{3}$(x - 2)

⇒ 3(y + 4) = 2(x - 2)

⇒ 3y + 12 = 2x - 4

⇒ 3y - 2x + 12 + 4 = 0

⇒ 3y - 2x + 16 = 0

Hence, equation of line is 3y - 2x + 16 = 0.

(iii) Since the line AB passes through the point P(k + 3, k - 4), the coordinates of P must satisfy the equation of the line, x - 2y - 10 = 0.

⇒ 3(k - 4) - 2(k + 3) + 16 = 0

⇒ 3k - 12 - 2k - 6 + 16 = 0

⇒ (3k - 2k) + (-12 - 6 + 16) = 0

⇒ k - 2 = 0

⇒ k = 2.

Hence, k = 2.