Show on a diagram the position of the straight line `x"cos"30^(@)+y"sin"30^(@)=2` in relation to the co-ordinate axes, indicating clearly which angle is `30^(@)` and which length is 2 units. Find
(i) the equation of the straight line parallel to that given line and passing through the point (4, 3) and
(ii) the length of the perpendicular from the origin on to this line

Answer

To solve the problem step by step, we will first analyze the given equation of the line, plot it, and then find the required parallel line and the distance from the origin.

### Step 1: Analyze the given line equation
The equation of the line is given as:
\[ x \cos 30^\circ + y \sin 30^\circ = 2 \]

Using the values of \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\), we can rewrite the equation:
\[ x \cdot \frac{\sqrt{3}}{2} + y \cdot \frac{1}{2} = 2 \]
Multiplying through by 2 to eliminate the fractions:
\[ \sqrt{3}x + y = 4 \]

### Step 2: Find intercepts to plot the line
To plot the line, we need to find the x-intercept and y-intercept.

1. **Finding the y-intercept** (set \(x = 0\)):
\[ \sqrt{3}(0) + y = 4 \implies y = 4 \]
So, the y-intercept is \((0, 4)\).

2. **Finding the x-intercept** (set \(y = 0\)):
\[ \sqrt{3}x + 0 = 4 \implies x = \frac{4}{\sqrt{3}} \approx 2.31 \]
So, the x-intercept is \(\left(\frac{4}{\sqrt{3}}, 0\right)\).

### Step 3: Plot the line
On a coordinate plane, plot the points:
- A (0, 4) on the y-axis
- B \(\left(\frac{4}{\sqrt{3}}, 0\right)\) on the x-axis

Draw a straight line through these two points. Mark the angle \(30^\circ\) between the x-axis and the line perpendicular to the given line (the normal), and indicate that the distance from the origin to the line is 2 units.

### Step 4: Find the equation of the parallel line through (4, 3)
Since the slope of the line \(\sqrt{3}x + y = 4\) is \(-\sqrt{3}\) (from the rearranged form \(y = -\sqrt{3}x + 4\)), a line parallel to it will have the same slope.

Using the point-slope form of the line:
\[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1) = (4, 3)\) and \(m = -\sqrt{3}\):
\[ y - 3 = -\sqrt{3}(x - 4) \]
Expanding this:
\[ y - 3 = -\sqrt{3}x + 4\sqrt{3} \]
Rearranging gives:
\[ y + \sqrt{3}x - (4\sqrt{3} + 3) = 0 \]

### Step 5: Find the distance from the origin to the parallel line
The distance \(D\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line:
- \(A = \sqrt{3}\)
- \(B = 1\)
- \(C = -(4\sqrt{3} + 3)\)
And the point is the origin \((0, 0)\):
\[ D = \frac{|\sqrt{3}(0) + 1(0) - (4\sqrt{3} + 3)|}{\sqrt{(\sqrt{3})^2 + (1)^2}} = \frac{|-(4\sqrt{3} + 3)|}{\sqrt{3 + 1}} = \frac{4\sqrt{3} + 3}{2} \]

### Final Answers:
1. The equation of the parallel line is:
\[ y + \sqrt{3}x - (4\sqrt{3} + 3) = 0 \]
2. The length of the perpendicular from the origin to this line is:
\[ D = \frac{4\sqrt{3} + 3}{2} \]