IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry - co-ordinate geometry

sachin_gupta_1991

Find the area of the rhombus enclosed by the lines ax+by+c=0 and ax-by-c=0.

waterdemon

Dude I will answer the question definitely but

I think the question must have been that

Find the area of the Rhombus anclosed in the lines

ax

by

c = 0

This becomes $ lines that will now act as 4 Sides for the Rhombus.

vivek54761

you r right waterdemon

waterdemon

For finding the Area first we need to find the co-ordinates of the sides of the Rhombus

So

We get four equations of lines from ax

bx

c = 0

1) ax + bx + c = 0

2) ax + by - c = 0

3) ax - by + c = 0

4) ax - by - c = 0

To get x and y coordinates of the lines.

*First lets take eqution 1

Putting x=0 we will get

y = -c / b

Putting y=0 we will get

x=-c / a

*Similarly we will get the coordinates of the remaining 3 lines, Now the coordinates of the four lines together will be as Follows :-

1) (-c / a , -c / b)

2)( +c / a,+c / b)

3)(-c / a ,+c / b)

4)( +c / a ,-c / b)

Now plot them on graph and u will get the diagonal of the Rhombus along the

X - axis and Y-axis

The length of the diagonal will be "2c/a"

Now applying the area of Rhombus formula

and get the answer

plZ rate me for my efforts

Cheers !!!!!!!!!!!!!!!!!!!!!