Th points A, B and C are (4, 0), (2, 2) and (0, 6) respectively. Find the equations of AB and BC.

If AB cuts the y-axis at P and BC cuts the x-axis at Q, find the coordinates of P and Q.

By formula,

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

Substituting values we get,

$\text{Slope of AB} = \dfrac{2 - 0}{2 - 4} \\[1em] = \dfrac{2}{-2} \\[1em] = -1.$

By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = -1(x - 4)

⇒ y = -x + 4

⇒ x + y = 4.

$\text{Slope of BC} = \dfrac{6 - 2}{0 - 2} \\[1em] = \dfrac{4}{-2} \\[1em] = -2.$

By point-slope form,

Equation of BC is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = -2(x - 2)

⇒ y - 2 = -2x + 4

⇒ 2x + y = 6.

Let point P be (0, a) and Q be (b, 0).

Substituting value of P in equation of AB we get,

⇒ 0 + a = 4

⇒ a = 4.

Substituting value of Q in equation of BC we get,

⇒ 2b + 0 = 6

⇒ 2b = 6

⇒ b = 3.

Hence, equation of AB is x + y = 4 and BC is 2x + y = 6 and P = (0, 4) and Q = (3, 0).