Question

If the area of parallelogram formed by the lines 2x-3y+a=0 , 3x-2y-a=0 , 2x-3y+3a=0 , and 3x-2y-2a=0 is 10 sq. Unit , then a is equal to

Answer 1

there is a direct frmula to solve this problem.

Area of paralellogram formed by the lines a1x+b1y+c1=0,a2x+b2y+d1=0,a1x+b1y+c2=0 and a2x+b2y+d2=0 is given by:

Answer 2

Concept

The area of a parallelogram is the space enclosed within its four sides. Area is equal to the product of length and height of the parallelogram.

Given

Equations of lines of parallelogram: 2x-3y+a=0,3x-2y-a=0,2x-3y+3a=0 and 3x-2y-2a=0.

Area of parallelogram=10 square units.

To find

The value of a when the area of parallelogram is 10 square units.

Explanation

The area of parallelogram is equal to Base* Height.

The coefficients of x and y are same and only fixed element is different.

Area= (3a-a){-2a-(-a)} divided by the determinant of the matrix $\left[\begin{array}{ccc}2&-3\\3&-2\\\end{array}\right]$.

10=I  2a*(-a)/(-4+9) I

10=2$a^{2}$/5

50=2$a^{2}$

$a^{2}$=25

a=-5 and a=5.

Hence if the area of the parallelogram is 10 then the value of a is 5 and -5.

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