If  two of the adjacent sides of a rectangle have equations x+3y = 8 and ax + y = 4, how can you find a?  

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

The two adjacent sides of a rectangle  are perpendicular.

The  product of the slopes of two straight lines  should be -1. These properties are sufficient to determine the a  in the equations  x+3y = 8 and ax + y = 4 which represent two adkacent sides of a rectangle.

Slope of the line, ax^2+bx+c = 0 is -a/b.

Therefore the slope of the line  x+3y = 8  is -1/3 and  the line ax+ + y = 4 is  -a/1 = -a.

Threfore the product of the slopes = -1. Or

(-1/3)(-a) = -1.

a/3 = -1.

a = -3.

Therefore the second equation , ax+y = 4 should be -3x+y =4

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

We know that the adjacent sides of a rectangle are perpendicular to each other. Now if a line has a slope s, the slope of the line perpendicular to it is -1/s.

Here we’ll use the relation that for a line ax+ by +c= 0 the slope is given by (–a/b)

Let’s take the lines given to us:

For x+3y = 8, the slope is -1/3

Therefore the slope of ax+ y = 4 should be 3

=> -a/1 = 3

=> a = -3

Therefore a is equal to -3.

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