IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

ashish_banga

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

Aatish

my answer is.......... $\frac{1}{m-n}$ .........

posting my full explanation soon........

ashish_banga

yes please tell the method

Aatish

Given equations are..........

----------------------(1)

-----------------(2)

-----------------------(3)

--------------------(4)

solving in pairs,,,,,,,(1) & (3),,,,,,(1) & (4),,,,,,,(2)&(4).........

we get three vertices......... $(\frac{1}{m-n},\frac{m}{m-n})$ ....(0,1) and (0,0)........

finding the area of the triangle formed by them.....using this determinant

$Area = \frac{1}{2}\begin{vmatrix}{x}_{1} & {y}_{1} & 1\\ {x}_{2} & {y}_{2} & 1\\ {x}_{3} & {y}_{3} & 1\end{vmatrix}$

multiply it with two.........to get the area of parallelogram..........

studyid

determine the perpendicular distance between the parallel limes y = mx and y = mx + 1

and determine the inersection sets (i.e. the coordinates of the points of intersection) of lines y = nx +1 with y = mx and y = nx with y = mx .

u get two points as (0,0) and [(1 / (m-n)), (m / (m-n))]

determine the distance between the two points

and then the area is given by

base * altitude :)

and u will get the required answer :)

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