If the point `x_1+t(x_2-x_1),y_1+t(y_2-y_1)` divides the join of `(x_1,y_1)` and `(x_2,y_2)` internally in ratio of `t:1` then vlaue of `t` is

Options

A. `t lt 0`
B. `0 lt t lt 1`
C. `t gt 1`
D. `t=1`

Explanation

To solve the problem, we need to find the value of \( t \) given that the point \( (x_1 + t(x_2 - x_1), y_1 + t(y_2 - y_1)) \) divides the line segment joining the points \( (x_1, y_1) \) and \( (x_2, y_2) \) internally in the ratio \( t:1 \).

### Step-by-Step Solution:

1. **Understanding the Section Formula**:
The coordinates of a point \( P \) that divides the line segment joining points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) in the ratio \( m:n \) are given by:
\[ P\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \]
In our case, \( m = t \) and \( n = 1 \).

2. **Applying the Section Formula**:
Using the section formula, the coordinates of point \( P \) can be expressed as:
\[ P\left( \frac{tx_2 + x_1}{t + 1}, \frac{ty_2 + y_1}{t + 1} \right) \]

3. **Equating Coordinates**:
We know that the coordinates of point \( P \) are also given as \( (x_1 + t(x_2 - x_1), y_1 + t(y_2 - y_1)) \). Therefore, we can set up the following equations:
\[ \frac{tx_2 + x_1}{t + 1} = x_1 + t(x_2 - x_1) \]
\[ \frac{ty_2 + y_1}{t + 1} = y_1 + t(y_2 - y_1) \]

4. **Solving the x-coordinate Equation**:
Let's solve the first equation:
\[ \frac{tx_2 + x_1}{t + 1} = x_1 + t(x_2 - x_1) \]
Cross-multiplying gives:
\[ tx_2 + x_1 = (x_1 + t(x_2 - x_1))(t + 1) \]
Expanding the right-hand side:
\[ tx_2 + x_1 = x_1(t + 1) + t(x_2 - x_1)(t + 1) \]
Simplifying further:
\[ tx_2 + x_1 = tx_1 + x_1 + t^2(x_2 - x_1) + t(x_2 - x_1) \]
Cancel \( x_1 \) from both sides:
\[ tx_2 = tx_1 + t^2(x_2 - x_1) + t(x_2 - x_1) \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ tx_2 - tx_1 = t^2(x_2 - x_1) + t(x_2 - x_1) \]
Factoring out \( t \):
\[ t(x_2 - x_1) = t(x_2 - x_1)(t + 1) \]

6. **Dividing Both Sides**:
If \( x_2 \neq x_1 \), we can divide both sides by \( (x_2 - x_1) \):
\[ t = t(t + 1) \]

7. **Setting Up the Quadratic Equation**:
Rearranging gives:
\[ t^2 + t - t = 0 \implies t^2 = 0 \]

8. **Finding the Value of t**:
Thus, we find:
\[ t = 0 \]

### Final Answer:
The value of \( t \) is \( 0 \).