Step-by-step explanation:
To find \( 2l + m \) where \( (I, m) \) are the coordinates of point \( S \), follow these steps:
1. **Determine Coordinates of Point Q:**
Given that \( Q \) lies on the y-axis, its coordinates are \( (0, y) \).
2. **Use the Right Angle Information:**
Since \( \angle PQR = 90^\circ \), vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{QR} \) are perpendicular.
- Vector \( \overrightarrow{PQ} = (0 - 0, y - 1) = (0, y - 1) \)
- Vector \( \overrightarrow{QR} = (4 - 0, 5 - y) = (4, 5 - y) \)
The dot product of \( \overrightarrow{PQ} \) and \( \overrightarrow{QR} \) should be zero:
\[
\overrightarrow{PQ} \cdot \overrightarrow{QR} = 0 \times 4 + (y - 1)(5 - y) = 0
\]
Simplify:
\[
(y - 1)(5 - y) = 0
\]
This gives two solutions:
\[
y - 1 = 0 \quad \text{or} \quad 5 - y = 0
\]
\[
y = 1 \quad \text{or} \quad y = 5
\]
- If \( y = 1 \), point \( Q \) is \( (0, 1) \), which is the same as \( P \), so this is not possible.
- Therefore, \( y = 5 \). So, \( Q \) is \( (0, 5) \).
3. **Find Coordinates of Point S:**
The line \( PS \) is a segment of the line \( 2y + x = 2 \). Point \( P \) is \( (0, 1) \), so substitute \( P \) into the line equation to get the line:
\[
2 \times 1 + x = 2
\]
\[
x = 0
\]
Hence, \( P \) is on the line \( 2y + x = 2 \). Now find the coordinates of \( S \) by ensuring that \( S \) is also on this line. Let \( S \) be \( (I, m) \):
\[
2m + I = 2
\]
4. **Find Coordinates Using Kite Properties:**
In a kite, diagonals are perpendicular. Therefore, diagonals \( PR \) and \( QS \) intersect at right angles.
- Diagonal \( PR \) can be computed:
\[
\text{Slope of } PR = \frac{5 - 1}{4 - 0} = 1
\]
Hence, the line \( PR \) is \( y - 1 = 1(x - 0) \), i.e., \( y = x + 1 \).
- The diagonal \( QS \) is perpendicular to \( PR \) and should have a slope of \( -1 \) (negative reciprocal):
\[
\text{Slope of } QS = -1
\]
Given \( Q = (0, 5) \), the line equation is:
\[
y - 5 = -1(x - 0)
\]
\[
y = -x + 5
\]
5. **Find Intersection of Diagonals:**
Solve the equations:
\[
y = x + 1
\]
\[
y = -x + 5
\]
Set them equal:
\[
x + 1 = -x + 5
\]
\[
2x = 4
\]
\[
x = 2
\]
Substitute \( x = 2 \) into \( y = x + 1 \):
\[
y = 2 + 1 = 3
\]
So, \( S = (2, 3) \).
6. **Calculate \( 2l + m \):**
The coordinates of \( S \) are \( (I, m) = (2, 3) \):
\[
2l + m = 2 \times 2 + 3 = 4 + 3 = 7
\]
Hence, \( 2l + m = 7 \).