# If 4x+ay=0, 5x+2y=0, bx+cy=0 and dx+4y=0 are the sides of a square find a, b, c and d.

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We pressume that

4x+ay = 0..........(1)

5x+2y = 0...........(2)

bx+cy = 0...........(3)

dx+4y = 0 .........(4)

are the sides of a square.

Since there is no constant terms in any of the 4 equations, all the 4 equations pass through the origin.

So the square is of zero length and zero width.

But yet we consider the lines at (1) and (2) are perpendicular. similarly lnes at (2) and (3)

So product of their slopes should be -1.So (-a/4)(-2/5) = -1. Or a = 20/-2 = **-10.**

Similarly from (2) and(3): (-2/5)(-c)/b) = -1 . Or c/b = -5/2

Similarly from the perpendicular lines (3) and (4) we get:

-c/b = d/4.

- -5/2 = d/4. Or d = 10.

Therefore the equations are:

4x-10y = 0

5x+2y = 0

2x--5y = 0

10x+4y = 0

The values of constants: a = -10 , b = 2 , c = - 5 ang d = 10.

In the question you have given the equations of the sides as 4x+ay=0, 5x+2y=0, bx+cy=0 and dx+4y=0. I am going to take these as the equations of adjoining sides.

Now the opposite sides of a square are parallel to each other and the adjoining sides are perpendicular to each other.

Let's convert all the lines to the slope and y- intercept form:

4x+ay=0 => y= (-4 /a) x ....(1)

5x+2y=0 => y= (-5/2 ) x ....(2)

bx+cy=0 => y = (-b/c) x ....(3)

and dx+4y=0 => y = (-d/4) x ....(4)

Now, slope of 1 = slope of 3,

=> -4/a = -b/c

=> a= 4c/b

slope of 2 = slope of 4

=> -5/2 = -d/4

=> d = 5*4/2= 10

Also, 1 and 2 are perpendicular, so (-4 /a) = -1/ (-5/2 )

=> a/4 = -5/2

=> a =-5*4/2 = -10

And 3 and 4 are perpendicular, so –b/c= -1/ (-d/4)

=> c/b = -d/4

=> d= -4*c/b

From both a= 4c/b= -10 and d = -4 *c/b =10 we can only get that c = -10b/4 = -5b/2

**Therefore using the given equations we can only determine d=10 and a=-10. **

**b and c can take any value such that c= -5*b/2.**