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