Gurveer and Arushi built a robot that can paint a path as it moves on a graph paper. Some co-ordinate of points are marked on it. It starts from (0, 0), moves to the points listed in order (in straight lines) and ends at (0, 0).

Arushi entered the points P(8, 6), Q(12, 2) and S(- 6, 6) in order The path drawn by robot is shown in the figure.

Based on the above, answer the following questions :

(i) Determine the distance OP.

ii) QS is represented by equation 2x + 9y = 42. Find the co-ordinates of the point where it intersects y - axis.

iii) (a) Point R(4.8, y) divides the line segment OP in a certain ratio. Find the ratio. Hence, find the value of y

OR

(iii) (b) Using distance formula, show that $\frac{PQ}{OS} = \frac{2}{3}$

$= \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$

So, the distance $OP$ is $10$ units.

(ii) $QS$ is represented by the equation $2x + 9y = 42$. Find the coordinates of the point where it intersects the y-axis.

To find the intersection of the line with the y-axis, we set $x = 0$ because any point on the y-axis has the form $(0, y)$. Substituting $x = 0$ into the equation of the line $2x + 9y = 42$, we get:

$2(0) + 9y = 42$

$9y = 42$

$y = \frac{42}{9} = 4.67$

Thus, the coordinates of the point where the line intersects the y-axis are $(0, 4.67)$.

(iii) (a) Point $R(4.8, y)$ divides the line segment $OP$ in a certain ratio. Find the ratio. Hence, find the value of $y$.

The coordinates of $O$ are $(0, 0)$ and the coordinates of $P$ are $(8, 6)$. Point $R$ divides the line segment $OP$ in some ratio. Using the section formula, if a point $(x, y)$ divides a line segment joining two points $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m:n$, then the coordinates of the point are given by:

$x = \frac{mx_2 + nx_1}{m+n}$

$y = \frac{my_2 + ny_1}{m+n}$

For point $R(4.8, y)$, we know the $x$-coordinate of $R$ is $4.8$. Using the section formula for the $x$-coordinate:

$4.8 = \frac{m \cdot 8 + n \cdot 0}{m + n}$

$4.8 = \frac{8m}{m + n}$

Multiplying both sides by $(m + n)$:

$4.8(m + n) = 8m$

$4.8m + 4.8n = 8m$

$4.8n = 3.2m$

$n = \frac{3.2}{4.8}m = \frac{2}{3}m$

So, the ratio is $m:n = 3:2$.

Now, we can use the $y$-coordinate of point $R$ to find the value of $y$. Using the section formula for the $y$-coordinate:

$y = \frac{m \cdot 6 + n \cdot 0}{m + n} = \frac{6m}{m + n}$

Substitute $n = \frac{2}{3}m$ into the equation:

$y = \frac{6m}{m + \frac{2}{3}m} $

$= \frac{6m}{\frac{5}{3}m} $

$= \frac{6 \cdot 3}{5} = \frac{18}{5} = 3.6$

Thus, the value of $y$ is $3.6$.

(iii) (b) Using the distance formula, show that $\frac{PQ}{OS} = \frac{2}{3}$.

First, let's find the distances $PQ$ and $OS$.

The coordinates of $P$ are $(8, 6)$ and the coordinates of $Q$ are $(12, 2)$. The distance $PQ$ is:

$PQ = \sqrt{(12 - 8)^2 + (2 - 6)^2} $

$= \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} $

$= 4\sqrt{2}$

The coordinates of $O$ are $(0, 0)$ and the coordinates of $S$ are $(-6, 6)$. The distance $OS$ is:

$OS = \sqrt{(-6 - 0)^2 + (6 - 0)^2} $

$= \sqrt{(-6)^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} $

$= 6\sqrt{2}$

Now, calculate the ratio $\frac{PQ}{OS}$:

$\frac{PQ}{OS} = \frac{4\sqrt{2}}{6\sqrt{2}} = \frac{4}{6} = \frac{2}{3}$

Thus, $\frac{PQ}{OS} = \frac{2}{3}$.