Find and write down a proof that the product of the gradients of two perpendicular lines is -1
Use trigonometry.
Let the slope of a line from A to B be m1. Let the slope of a perpendicular line from A to C be m2. The tangent (or slope) of the line between these two lines is:
tan (a - b) = (m1 + m2)/(1 - m1 m2)
a and b are the ANGLES of the two lines, measured from the x axis.
(This is a trigonometric identity)
Now IF m1 m2 = -1, then
tan (a-b) = (m1 - 1/m1)/0 = infinity
therefore the angle between the two lines is 90 degrees.
To prove that the product of the gradients (slopes) of two perpendicular lines is -1, we can use the concept of trigonometry.
Suppose we have two perpendicular lines, Line 1 and Line 2. Let the gradients (slopes) of Line 1 and Line 2 be denoted by m1 and m2, respectively.
The slope of a line can be expressed as the tangent of the angle it makes with the positive x-axis. Let α1 and α2 be the angles that Line 1 and Line 2 make with the positive x-axis, respectively.
Using trigonometric identities, we have tan(α1) = m1 and tan(α2) = m2.
Now, to find the angle between two lines, we can use the difference of angles formula for tangent:
tan(α1 - α2) = (tan(α1) - tan(α2))/(1 + tan(α1)tan(α2))
Since α1 and α2 are right angles (perpendicular lines), their difference α1 - α2 is 90 degrees or π/2 radians.
Substituting the given expressions for tan(α1) = m1 and tan(α2) = m2, we have:
tan(π/2) = (m1 - m2)/(1 + m1m2)
The tangent of π/2 is undefined, which means the denominator (1 + m1m2) must be zero:
1 + m1m2 = 0
Rearranging the equation, we get:
m1m2 = -1
Thus, we have proved that the product of the gradients (slopes) of two perpendicular lines is -1.