# 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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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 **a**x + 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 **a**x+ + 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

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.**