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To solve these problems, we use the formula to find the perpendicular distance from a point \((x_1, y_1)\) to a line given by \(Ax + By + C = 0\):

\[
d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]

Let's solve each question step by step.

### Problem 28:
**Line equation:** \(4x - 6y - 3 = 0\)

**Point:** \((2, 3)\)

Here, \(A = 4\), \(B = -6\), and \(C = -3\). The point is \((x_1, y_1) = (2, 3)\).

The distance is:

\[
d = \frac{|4(2) - 6(3) - 3|}{\sqrt{4^2 + (-6)^2}} = \frac{|8 - 18 - 3|}{\sqrt{16 + 36}} = \frac{|8 - 21|}{\sqrt{52}} = \frac{13}{\sqrt{52}} = \frac{13}{2\sqrt{13}} = \frac{\sqrt{13}}{2}
\]

So, the length of the perpendicular from the point \((2, 3)\) to the line is:

\[
\frac{\sqrt{13}}{2}
\]

### Problem 29:
**Line equation:** \(2x + 3y - 6 = 0\)

**Point:** \((2, 5)\)

Here, \(A = 2\), \(B = 3\), and \(C = -6\). The point is \((x_1, y_1) = (2, 5)\).

The distance is:

\[
d = \frac{|2(2) + 3(5) - 6|}{\sqrt{2^2 + 3^2}} = \frac{|4 + 15 - 6|}{\sqrt{4 + 9}} = \frac{|13|}{\sqrt{13}} = \frac{13}{\sqrt{13}} = \sqrt{13}
\]

So, the length of the perpendicular from the point \((2, 5)\) to the line is:

\[
\sqrt{13}
\]

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Would you like any further clarifications or details?

### Related Questions:
1. How do you find the equation of a line perpendicular to a given line?
2. What is the geometric interpretation of the perpendicular distance between a point and a line?
3. How does the perpendicular distance formula apply to 3D space?
4. Can we use vectors to find the perpendicular distance from a point to a line?
5. How does the slope of a line relate to its perpendicular distance?

### Tip:
For quick calculations of perpendicular distances, remember to simplify the line equation into the general form \(Ax + By + C = 0\) before applying the formula.

Find the length of perpendicular from the point (2,3) on the line 4x-6y-3=0.
Find the length of perpendicular from the point (2,5) on the line 2x+3y-6=0.

Learn how to calculate the perpendicular distance from a point to a line using the formula d = |Ax1 + By1 + C| / sqrt(A^2 + B^2). This problem covers key concepts in coordinate geometry and is suitable for students in Grades 10-12.

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