In the given figure, ABC is a triangle and BC is parallel to the y-axis. AB and AC intersect the y-axis at P and Q respectively. Find :

(i) the co-ordinates of A

(ii) the length of AB and AC

(iii) the ratio in which Q divides AC

(iv) the equation of the line AC

(i) From figure,

The co-ordinates of A = (4, 0).

(ii) By distance formula,

D = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Substituting values we get,

$AB = \sqrt{(-2 - 4)^2 + (3 - 0)^2} \\[1em] = \sqrt{(-6)^2 + (3)^2} \\[1em] = \sqrt{36 + 9} \\[1em] = \sqrt{45} \\[1em] = 3\sqrt{5}. \\[1em] AC = \sqrt{(-2 - 4)^2 + (-4 - 0)^2} \\[1em] = \sqrt{(-6)^2 + (-4)^2} \\[1em] = \sqrt{36 + 16} \\[1em] = \sqrt{52} \\[1em] = 2\sqrt{13}.$

Hence, length of $AB = 3\sqrt{5}\text{ and } AC = 2\sqrt{13}$.

(iii) From figure,

Q lies on y-axis.

∴ x co-ordinate of Q = 0.

Let co-ordinate of Q are (0, a).

Let ratio in which Q divides AC be k : 1.

By section-formula,

$Q = \Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Comparing x-coordinate we get :

$\Rightarrow 0 = \dfrac{k \times -2 + 1 \times 4}{k + 1} \\[1em] \Rightarrow 0 = -2k + 4 \\[1em] \Rightarrow 2k = 4 \\[1em] \Rightarrow k = 2.$

k : 1 = 2 : 1.

Hence, Q divides AC in the ratio 2 : 1.

(iv) By formula,

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

Slope of AC = $\dfrac{-4 - 0}{-2 - 4} = \dfrac{-4}{-6} = \dfrac{2}{3}$

By point-slope form,

Equation of AC is :

⇒ y - y₁ = m (x - x₁)

⇒ y - 0 = $\dfrac{2}{3}$ (x - 4)

⇒ 3y = 2x - 8

⇒ 2x - 3y = 8

Hence, equation of AC is 2x - 3y = 8.