If each of the points `A(x_(1),4),B(-2,y_(1))` lies on the line joining the points C(2,-1),D(5,-3) then the point `P(x_(1),y_(1))` lies on the line

Options

A. `6(x+y)-25=0`
B. `2x+6y+1=0`
C. `2x3y+16=0`
D. `6(x+y)-23=0`

Answer

To solve the problem, we need to find the coordinates of the points A and B that lie on the line joining points C and D. Then we will determine the coordinates of point P.

### Step-by-Step Solution:

1. **Identify the coordinates of points C and D:**
- C(2, -1)
- D(5, -3)

2. **Calculate the slope of line CD:**
The slope (m) of a line through points (x1, y1) and (x2, y2) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
For points C(2, -1) and D(5, -3):
\[ m = \frac{-3 - (-1)}{5 - 2} = \frac{-3 + 1}{5 - 2} = \frac{-2}{3} \]

3. **Write the equation of the line CD using point-slope form:**
The point-slope form of the line is given by:
\[ y - y_1 = m(x - x_1) \]
Using point C(2, -1):
\[ y - (-1) = -\frac{2}{3}(x - 2) \]
Simplifying this:
\[ y + 1 = -\frac{2}{3}x + \frac{4}{3} \]
\[ y = -\frac{2}{3}x + \frac{4}{3} - 1 \]
\[ y = -\frac{2}{3}x + \frac{4}{3} - \frac{3}{3} \]
\[ y = -\frac{2}{3}x + \frac{1}{3} \]

4. **Convert the equation to standard form:**
Multiply through by 3 to eliminate the fraction:
\[ 3y = -2x + 1 \]
Rearranging gives:
\[ 2x + 3y - 1 = 0 \]

5. **Substitute point A(x1, 4) into the line equation:**
Substitute \(y = 4\):
\[ 2x_1 + 3(4) - 1 = 0 \]
\[ 2x_1 + 12 - 1 = 0 \]
\[ 2x_1 + 11 = 0 \]
\[ 2x_1 = -11 \implies x_1 = -\frac{11}{2} \]

6. **Substitute point B(-2, y1) into the line equation:**
Substitute \(x = -2\):
\[ 2(-2) + 3y_1 - 1 = 0 \]
\[ -4 + 3y_1 - 1 = 0 \]
\[ 3y_1 - 5 = 0 \]
\[ 3y_1 = 5 \implies y_1 = \frac{5}{3} \]

7. **Conclusion:**
The coordinates of point P(x1, y1) are:
\[ P\left(-\frac{11}{2}, \frac{5}{3}\right) \]