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

(i) Write the co-ordinates of A.

(ii) Find the length of AB and AC.

(iii) Find the ratio in which Q divides AC.

(iv) Find the equation of the line AC.

(i) From graph,

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

(ii) By distance formula,

Distance = $\sqrt{(y$2 - y1)^2 + (x2 - x1)^2}

Substituting values we get,

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

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

(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) \\[1em] \text{For x co-ordinate} \\[1em] \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) Slope of AC = $\dfrac{-4 - 0}{-2 - 4} = \dfrac{-4}{-6} = \dfrac{2}{3}$.

By point-slope form,

Equation of AB 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.