The vertices of the triangle are P(2,1), Q(4,-1) and R(3,2). If through P and R lines parallel to opposite sides are drawn to intersect in S, then what is the area of PQRS?(a) 6(b) 4(c) 8(d) 12

To solve the problem, we will compute step by step.

Step 1: Equation of lines parallel to opposite sides
Side QR: The slope of QR is given by: Slope of QR = (y₂ - y₁) / (x₂ - x₁) = (2 - (-1)) / (3 - 4) = 3 / (-1) = -3.

The line through P (2, 1) and parallel to QR will have the same slope (-3). Using the point-slope form of the equation of a line: y - y₁ = m(x - x₁), y - 1 = -3(x - 2), y = -3x + 6 + 1, y = -3x + 7.
Thus, the equation of the line parallel to QR passing through P is: y = -3x + 7.

Side PQ: The slope of PQ is: Slope of PQ = (y₂ - y₁) / (x₂ - x₁) = (-1 - 1) / (4 - 2) = -2 / 2 = -1.

The line through R (3, 2) and parallel to PQ will have the same slope (-1). Using the point-slope form: y - y₁ = m(x - x₁), y - 2 = -1(x - 3), y = -x + 3 + 2, y = -x + 5.
Thus, the equation of the line parallel to PQ passing through R is: y = -x + 5.

Step 2: Finding the coordinates of S
The point S is the intersection of the lines:

y = -3x + 7,
y = -x + 5.
Equating the two equations: -3x + 7 = -x + 5, -3x + x = 5 - 7, -2x = -2, x = 1.

Substitute x = 1 into y = -x + 5: y = -(1) + 5 = 4.

Thus, the coordinates of S are (1, 4).

Step 3: Area of quadrilateral PQRS
The vertices of PQRS are: P(2, 1), Q(4, -1), R(3, 2), S(1, 4).

Using the formula for the area of a quadrilateral with vertices
(
𝑥
1
,
𝑦
1
)
,
(
𝑥
2
,
𝑦
2
)
,
(
𝑥
3
,
𝑦
3
)
,
(
𝑥
4
,
𝑦
4
)
(x
1
​
,y
1
​
),(x
2
​
,y
2
​
),(x
3
​
,y
3
​
),(x
4
​
,y
4
​
): Area = 0.5 * |x1y2 + x2y3 + x3y4 + x4y1 - (y1x2 + y2x3 + y3x4 + y4x1)|.

Substitute the coordinates: Area = 0.5 * |(2 * -1) + (4 * 2) + (3 * 4) + (1 * 1) - [(1 * 4) + (-1 * 3) + (2 * 1) + (4 * 2)]|, Area = 0.5 * |(-2) + 8 + 12 + 1 - [4 - 3 + 2 + 8]|, Area = 0.5 * |19 - 11|, Area = 0.5 * 8 = 4.

Final Answer:
The area of PQRS is 4.
Correct option: (b) 4.