The graph of the linear eqation 3 x + 4y = 24 is a straight line intersecting x- axis and y- axis at the points A and B respectively `P (2,0) and Q (0,(3)/(2))` are two points on the sides OA and OB respectively of ` Delta OAB` , whete O is the origin of the co-ordinate system. Given that AB = 10 cm then PQ =

Options

A. 20 cm
B. 2 . 5 cm
C. 40 cm
D. 5 cm

Answer

To solve the problem, we need to find the length of segment PQ given the triangle OAB where O is the origin, A is the intersection of the line with the x-axis, and B is the intersection with the y-axis.

### Step-by-step Solution:

1. **Find the points A and B**:
The equation of the line is given as \(3x + 4y = 24\).
- To find point A (intersection with the x-axis), set \(y = 0\):
\[ 3x + 4(0) = 24 \implies 3x = 24 \implies x = 8 \]
So, point A is \(A(8, 0)\).
- To find point B (intersection with the y-axis), set \(x = 0\):
\[ 3(0) + 4y = 24 \implies 4y = 24 \implies y = 6 \]
So, point B is \(B(0, 6)\).

2. **Identify points P and Q**:
- Point P is given as \(P(2, 0)\) on side OA.
- Point Q is given as \(Q(0, \frac{3}{2})\) on side OB.

3. **Calculate distances OP and OQ**:
- The distance \(OP\) from the origin O(0, 0) to P(2, 0) is:
\[ OP = 2 \text{ cm} \]
- The distance \(OQ\) from the origin O(0, 0) to Q(0, \frac{3}{2}) is:
\[ OQ = \frac{3}{2} \text{ cm} \]

4. **Use the Pythagorean theorem to find PQ**:
Since triangle OPQ is a right triangle, we can apply the Pythagorean theorem:
\[ PQ^2 = OP^2 + OQ^2 \]
Substituting the values we found:
\[ PQ^2 = (2)^2 + \left(\frac{3}{2}\right)^2 \]
\[ PQ^2 = 4 + \frac{9}{4} \]
To add these, convert 4 to a fraction:
\[ 4 = \frac{16}{4} \]
Now, add the fractions:
\[ PQ^2 = \frac{16}{4} + \frac{9}{4} = \frac{25}{4} \]

5. **Find PQ**:
Taking the square root of both sides:
\[ PQ = \sqrt{\frac{25}{4}} = \frac{5}{2} = 2.5 \text{ cm} \]

### Final Answer:
The length of segment PQ is \(2.5 \text{ cm}\).