The vertex of the right angle of a right angled triangle lies on the straight line `2x +y- 10 =0` and the two other vertices, at points `(2,-3)` and `(4,1)` then the area of triangle in sq. units is-

Options

A. `sqrt(10)`
B. 3
C. `(33)/(5)`
D. 11

Answer

To solve the problem step by step, we will find the area of the right-angled triangle formed by the given points and the vertex on the line.

### Step 1: Identify the points and the line
We have two points:
- A(2, -3)
- B(4, 1)

The vertex of the right angle, point C, lies on the line given by the equation:
\[ 2x + y - 10 = 0 \]

### Step 2: Express y in terms of x
From the line equation, we can express y in terms of x:
\[ y = 10 - 2x \]

### Step 3: Find the slopes of the segments
To find the coordinates of point C, we need to ensure that the segments AB and BC are perpendicular. The slope of line segment AB can be calculated as follows:
\[ \text{slope of AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-3)}{4 - 2} = \frac{4}{2} = 2 \]

Let the coordinates of point C be (x, y). The slope of segment BC is:
\[ \text{slope of BC} = \frac{y - 1}{x - 4} \]

### Step 4: Set up the perpendicularity condition
Since AB is perpendicular to BC, the product of their slopes should equal -1:
\[ \text{slope of AB} \times \text{slope of BC} = -1 \]
\[ 2 \times \frac{y - 1}{x - 4} = -1 \]
Substituting \( y = 10 - 2x \) into the equation:
\[ 2 \times \frac{(10 - 2x) - 1}{x - 4} = -1 \]
\[ 2 \times \frac{9 - 2x}{x - 4} = -1 \]

### Step 5: Solve for x
Cross-multiplying gives:
\[ 2(9 - 2x) = -1(x - 4) \]
\[ 18 - 4x = -x + 4 \]
Rearranging the equation:
\[ 18 - 4 = -x + 4x \]
\[ 14 = 3x \]
\[ x = \frac{14}{3} \]

### Step 6: Find y-coordinate
Substituting \( x = \frac{14}{3} \) back into the equation for y:
\[ y = 10 - 2\left(\frac{14}{3}\right) = 10 - \frac{28}{3} = \frac{30}{3} - \frac{28}{3} = \frac{2}{3} \]

Thus, the coordinates of point C are:
\[ C\left(\frac{14}{3}, \frac{2}{3}\right) \]

### Step 7: Calculate the area of the triangle
The area of triangle ABC can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ \text{Area} = \frac{1}{2} \left| 2(1 - \frac{2}{3}) + 4(\frac{2}{3} + 3) + \frac{14}{3}(-3 - 1) \right| \]
Calculating each term:
\[ = \frac{1}{2} \left| 2 \cdot \frac{1}{3} + 4 \cdot \frac{11}{3} + \frac{14}{3} \cdot (-4) \right| \]
\[ = \frac{1}{2} \left| \frac{2}{3} + \frac{44}{3} - \frac{56}{3} \right| \]
\[ = \frac{1}{2} \left| \frac{-10}{3} \right| = \frac{5}{3} \]

### Final Result
The area of the triangle is:
\[ \text{Area} = \frac{5}{3} \text{ square units} \]