If a square is formed by x + 3y +4 = 0, 2x + 6y + 7 = 0, 3x – y + 4 = 0 and 6x – y + a = 0, what is the value of a?  

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have the lines x + 3y +4 = 0, 2x + 6y + 7 = 0, 3x – y + 4 = 0 and 6x – y + a = 0.

Now x + 3y +4 = 0 and 2x + 6y + 7 = 0 are parallel lines.

x + 3y +4 = 0 => y = -x/3 – 4/3

and  2x + 6y + 7 = 0 => y = -x/3 – 7/6

Therefore the distance between the two lines is  1/6

Now the lines  3x – y + 4 = 0 and 6x – 2y + a = 0 are perpendicular to the first two lines.

3x – y + 4 = 0 => y = 3x + 4

6x – 2y + a = 0 => y = 3x + a/2

Therefore as we have to form a square the distance between  y = 3x + 4 and y = 3x + a/2 should also be 1/6

=> 4 – a/2 = 1/6

=> a = 23/3

or a/2 – 4 = 1/6

=> a/2 = 4 + 1/6

=> a = 25/3

Therefore a can be equal to 25/3 or 23/3.

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neela | High School Teacher | (Level 3) Valedictorian

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To determine a if  the square is formed by x + 3y +4 = 0, 2x + 6y + 7 = 0, 3x – y + 4 = 0 and 6x – y + a = 0, what is the value of a:

The lines x+3y+4 = 0 and  3x-y+4 are perpendicular as the their slope is -1/3 and 3 and the product  slope is -1.

Therefore the other pair of lines 2x+6y +7 = 0 and 6x-y +a = 0 should be perpendicular. The slope of them are -2/6 and 6/1. The product of the slopes = -(2/6)(6/1) = -2. For the lines to be perpendicular the product of the slopes must be equal to -1.

For a square to be formed  any two adjacent lines  equations   must be perpendicular. Therefore the given four lines does not form a square.

Correcting the equation 6x-y + a = 0 as 6x-2y +a = 0, we get the pars of opposite lines:

x+3y+4 = 0 || 2x+6y+7 = 0 , Or x+3y+3.5 = 0

3x-y +4 = 0 ||  6x-2y +a = 0. Or 3x-y +a/2.

The distance between first pair of parallel lines  = {-4 -(-7)}/(3^2+1^2) = 3/sqrt10....................(1)

The distant between the other pair of || lines = |4- a/2|/sqrt(3^2+1^2) = (4-a/2)/sqrt10...........(2).

For a square, the distance beween the opposite pairs of lines must be same.

Therefore 3/sqrt10  = (4-a/2)/sqrt10.

Or 3 = (4-a/2).

Or a/2 = 4-3 = 1

 a= 2.

Therefore  the value of a = 2 int he equation 6x-2y +a  = 0 in order that the system of equations,

 x + 3y +4 = 0, 2x + 6y + 7 = 0, 3x – y + 4 = 0 and 6x – 2y + a = 0 should form a square.

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