Question

Area of the parallelogram formed by the lines y = mx, y = mx + 1, y = nx and y = nx + 1 equals

(a) [(|m + n|)/(m – n)²]

(b) [2/(|m + n|)]

(c) [1/(m – n)]

(d) [1/(|m + n|)]

Answer 1

The correct option (c) [1/(m – n)]

Explanation:

y = mx, y = mx + 1, y = nx and y = nx + 1 from a parallelogram.

solving equation, vertices are 0(0, 0), A[{1/(m – n)}, y], B(0, 1) & C[{1/(n – m)}, y]

Now area of OABC  = 2 area of ΔOAB

= 2 × (1/2) × base × height 

= (1 – 0) × [1/(m – n)] 

= [1/(m – n)]

= [1/(|m – n|)]

because Two vertices O & B lie on y-axis. hence area is (1/2) × base × height

where height is abscissa of point A.

∴ area of OABC = [1/(|m – n|)]