Find the length of the perpendicular drawn from the point (4, - 3) upon the straight line 6x + 3y – 10=0, the angle between the axes being `60^@`.

Answer

To find the length of the perpendicular drawn from the point (4, -3) to the straight line given by the equation \(6x + 3y - 10 = 0\) with the axes inclined at an angle of \(60^\circ\), we can use the formula for the length of the perpendicular from a point to a line in an oblique coordinate system.

### Step-by-Step Solution:

1. **Identify the coefficients from the line equation:**
The line equation is \(6x + 3y - 10 = 0\).
Here, \(a = 6\), \(b = 3\), and \(c = -10\).

2. **Identify the coordinates of the point:**
The point from which we are drawing the perpendicular is \((h, k) = (4, -3)\).

3. **Use the formula for the length of the perpendicular:**
The formula for the length \(L\) of the perpendicular from a point \((h, k)\) to the line \(ax + by + c = 0\) in an oblique coordinate system is given by:
\[ L = \frac{|ah + bk + c|}{\sqrt{a^2 + b^2 - 2ab \cos \omega}} \]
where \(\omega\) is the angle between the axes, which is \(60^\circ\) in this case.

4. **Calculate the numerator:**
Substitute \(h\), \(k\), \(a\), \(b\), and \(c\) into the numerator:
\[ |ah + bk + c| = |6 \cdot 4 + 3 \cdot (-3) - 10| \]
\[ = |24 - 9 - 10| = |5| = 5 \]

5. **Calculate the denominator:**
First, calculate \(a^2 + b^2\):
\[ a^2 + b^2 = 6^2 + 3^2 = 36 + 9 = 45 \]
Next, calculate \(2ab \cos \omega\):
\[ 2ab \cos 60^\circ = 2 \cdot 6 \cdot 3 \cdot \frac{1}{2} = 18 \]
Now, substitute these values into the denominator:
\[ \sqrt{a^2 + b^2 - 2ab \cos \omega} = \sqrt{45 - 18} = \sqrt{27} = 3\sqrt{3} \]

6. **Combine the results to find \(L\):**
Now substitute the values into the formula for \(L\):
\[ L = \frac{5}{3\sqrt{3}} \]

7. **Rationalize the denominator:**
To rationalize the denominator, multiply the numerator and the denominator by \(\sqrt{3}\):
\[ L = \frac{5\sqrt{3}}{3 \cdot 3} = \frac{5\sqrt{3}}{9} \]

### Final Answer:
The length of the perpendicular from the point (4, -3) to the line \(6x + 3y - 10 = 0\) is \(\frac{5\sqrt{3}}{9}\) units.