Let D is a point on the line `l_(1):x+y=2=0 and S(3, 3)` is a fixed point. The line `l_(2)` is perpendicular to DS and passes through S. If M is another point on the line `l_(1)` (other than D), then the locus of the point of intersection of `l_(2)` and the angle bisector of the angle MDS is

Options

A. `(x+y-2)^(2)=2(x-3)^(2)+2(y-3)^(2)`
B. `(x+y-2)^(2)=(x-2)^(2)+(y-3)^(2)`
C. `(x+y-2)^(2)=((x-3)^(2)+(y-3)^(2))/(2)`
D. None of these

Answer

To solve the problem step by step, we will follow the logical deductions based on the given conditions.

### Step 1: Define the Points and Lines
Let \( D(x_1, y_1) \) be a point on the line \( l_1: x + y - 2 = 0 \). This means that \( x_1 + y_1 = 2 \).

Let \( S(3, 3) \) be the fixed point.

### Step 2: Determine the Equation of Line \( l_2 \)
The line \( l_2 \) is perpendicular to the line segment \( DS \) and passes through point \( S \).

The slope of line \( DS \) can be calculated as:
\[ \text{slope of } DS = \frac{y_1 - 3}{x_1 - 3} \]
Thus, the slope of line \( l_2 \) (being perpendicular) is:
\[ \text{slope of } l_2 = -\frac{x_1 - 3}{y_1 - 3} \]

Using point-slope form, the equation of line \( l_2 \) is:
\[ y - 3 = -\frac{x_1 - 3}{y_1 - 3}(x - 3) \]

### Step 3: Find Another Point \( M \) on Line \( l_1 \)
Let \( M(x_2, y_2) \) be another point on line \( l_1 \). Therefore, we have:
\[ x_2 + y_2 - 2 = 0 \implies y_2 = 2 - x_2 \]

### Step 4: Determine the Angle Bisector of \( \angle MDS \)
The angle bisector of \( \angle MDS \) can be determined using the angle bisector theorem. The coordinates of the intersection point of the angle bisector and line \( l_2 \) will be denoted as \( P(h, k) \).

### Step 5: Establish the Relationship Between \( P \), \( D \), and \( S \)
Since \( P \) is on the angle bisector of \( \angle MDS \), we can use the property of angle bisectors. The distances from point \( P \) to points \( D \) and \( S \) must maintain a ratio based on the lengths of segments \( MD \) and \( MS \).

### Step 6: Use Perpendicular Distances
Let \( PF \) be the perpendicular from point \( P \) to line \( l_1 \). Since \( P \) lies on the angle bisector, we have:
\[ PF = PS \]
Using the distance formula, we can express these distances in terms of \( h \) and \( k \).

### Step 7: Set Up the Equation
The perpendicular distance from point \( P(h, k) \) to line \( l_1 \) is given by:
\[ PF = \frac{|h + k - 2|}{\sqrt{1^2 + 1^2}} = \frac{|h + k - 2|}{\sqrt{2}} \]
The distance \( PS \) is:
\[ PS = \sqrt{(h - 3)^2 + (k - 3)^2} \]

### Step 8: Equate the Distances
Setting the two distances equal gives:
\[ \frac{|h + k - 2|}{\sqrt{2}} = \sqrt{(h - 3)^2 + (k - 3)^2} \]

### Step 9: Square Both Sides
Squaring both sides to eliminate the square root yields:
\[ \frac{(h + k - 2)^2}{2} = (h - 3)^2 + (k - 3)^2 \]

### Step 10: Simplify the Equation
Expanding both sides and simplifying will yield the locus of point \( P \).

### Step 11: Final Locus Equation
After simplification, we arrive at the final equation that represents the locus of point \( P \).