In the given figure, line AB meets y-axis at point A. Line through C(2, 10) and D intersects line AB at right angle at point P. Find :

(i) equation of line AB.

(ii) equation of line CD.

(iii) co-ordinates of point E and D.

(i) By formula,

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

Substituting values we get,

$\text{Slope of AB} = \dfrac{8 - 6}{-6 - 0} \\[1em] = \dfrac{2}{-6} \\[1em] = -\dfrac{1}{3}.$

By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 6 = $-\dfrac{1}{3}$(x - 0)

⇒ 3(y - 6) = -1(x)

⇒ 3y - 18 = -x

⇒ x + 3y = 18.

Hence, equation of line AB is x + 3y = 18.

(ii) From figure,

CD is perpendicular to AB.

Slope of AB (m₁) = $-\dfrac{1}{3}$

Let slope of CD be m₂.

⇒ m₁ × m₂ = -1

⇒ $-\dfrac{1}{3} \times m_2 = -1$

⇒ m₂ = 3.

Equation of CD is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 10 = 3(x - 2)

⇒ y - 10 = 3x - 6

⇒ 3x - y - 6 + 10 = 0

⇒ 3x - y + 4 = 0.

Hence, equation of CD is 3x - y + 4 = 0.

(iii) From figure,

E lies on x-axis. Let co-ordinates of E be (a, 0).

Since, E lies on line AB it will satisfy its equation.

Substituting value of E in AB we get,

⇒ a + 3(0) = 18

⇒ a = 18.

E = (18, 0)

D lies on x-axis. Let co-ordinates of D be (b, 0).

Since, D lies on CD it will satisfy its equation.

Substituting value of D in CD we get,

3b - 0 + 4 = 0

3b = -4

b = $-\dfrac{4}{3}$.

D = $(-\dfrac{4}{3}, 0)$.

Hence, co-ordinates of E = (18, 0) and D = $(-\dfrac{4}{3}, 0)$.