What is the area of the figure bounded by lines described by the follo

Line A: 2x - y = 2

Line B: \(x^3 = 27\) or x = 3

Line C: 2x - y = -5

Line D: x = -2


We see that lines A and C have the same coefficient of x and y, and hence are parallel and line B and D are also parallel.

We construct lines A and C by joining the x intercept (putting y = 0) and the y intercept (putting x = 0)


Let P = Intersection point of A and B: We have x = 3 and putting x = 3 in 2x - y = 2, we get y = 4. P(3, 4)


Let Q = Intersection point of A and D: We have x = -2 and putting x = -2 in 2x - y = 2, we get y = -6. Q(-2, -6)


Let R = Intersection point of C and D: We have x = -2 and putting x = -2 in 2x - y = -5, we get y = 1. Q(-2, 1)


Let S = Intersection point of C and B: We have x = 3 and putting x = 3 in 2x - y = -5, we get y = 11. P(3, 11)



The 4 points are given as shown in the diagram below.

Using the distance formula between 2 points \((x_1,y_1) , (x_2,y_2)\) is given by

\(Distance = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }\)

PQ = \(\sqrt{(-2 – 3)^2 + (-6 – 4)^2 }= \sqrt{25 + 100} = \sqrt{125}\)

SR = \(\sqrt{(-2 – 3)^2 + (1 – 11)^2 }= \sqrt{25 + 100} = \sqrt{125}\)

RQ = \(\sqrt{(-2 – (-2))^2 + (1 – (-6))^2 }= \sqrt{0 + 49} = 7\)

SP = \(\sqrt{(3 - 3)^2 + (11 – 4)^2 }= \sqrt{0 + 49} = 7\)


The figure is either a parallelogram. It cannot be a rectangle because vertex angles of \(90^o\) cannot be formed by lines have a slope with lines parallel to the y axis.


Area = Base x Height.

Taking the base as PQ = 7 and the height as LM (=RT) = 5, The area is 7 * 5 = 35



Option D

Arun Kumar