prove that the length of the perpendicular from the origin to the straight line joining the two points having coordinates ( a cos alpha, a sin alpha) and (a cos beta, a sin beta) is
a cos { ( alpha + beta)/2}.
both points are on a circle of radius a.
The distance to the line is the altitude of the triangle formed by the two points and the circle center.
Draw the diagram and see the answer checks.
To prove the length of the perpendicular from the origin to the straight line joining two points, we can use the concept of dot product between two vectors.
Let's start by finding the direction vector of the line joining the two given points.
The first point has coordinates (a cos α, a sin α) and the second point has coordinates (a cos β, a sin β).
The direction vector of the line is obtained by subtracting the coordinates of the first point from the coordinates of the second point:
Direction vector = (a cos β - a cos α, a sin β - a sin α)
Now, consider the vector representing the length of the perpendicular from the origin to the line. Let's call this vector P.
Since P is perpendicular to the line, it should be orthogonal to the direction vector of the line.
Using the dot product, we can say that:
P · Direction vector = 0
Expanding this dot product:
P · (a cos β - a cos α, a sin β - a sin α) = 0
Now, let's find the expression for the vector P using the coordinates of the perpendicular.
The equation of a line is given by:
Ax + By + C = 0
Where A, B, and C are constants that depend on the coordinates of the points the line passes through.
The equation of the perpendicular can be written as:
x cos θ + y sin θ = 0
Where θ is the angle that the perpendicular makes with the positive x-axis.
We know that the coordinates of the points on the line satisfy this equation. Substituting the coordinates of the points, we have:
(a cos α) cos θ + (a sin α) sin θ = 0 (Point 1)
(a cos β) cos θ + (a sin β) sin θ = 0 (Point 2)
Now, let's solve these equations for cos θ and sin θ.
By subtracting equations (Point 2) - (Point 1), we have:
(a cos β - a cos α) cos θ + (a sin β - a sin α) sin θ = 0
Dividing both sides by a, we get:
(cos β - cos α) cos θ + (sin β - sin α) sin θ = 0
Using the identity cos(A - B) = cos A cos B + sin A sin B, we can rewrite this equation as:
cos(β - α) cos θ = sin(β - α) sin θ
Dividing both sides by cos(β - α) sin(θ) and sin(β - α) sin(θ), respectively, we have:
cot(θ) = tan(β - α)
Since cot(θ) = 1/tan(θ), we get:
tan(θ) = 1/tan(β - α)
This implies that θ = (β - α + π/2), where π/2 is added to account for the fact that the perpendicular lies in quadrant 2 when the angle (β - α) is in quadrant 1.
Now, we can substitute this value for θ back into the equation of the perpendicular:
x cos((β - α) + π/2) + y sin((β - α) + π/2) = 0
Expanding this equation further, we have:
x cos(β - α) cos(π/2) - x sin(β - α) sin(π/2) + y sin(β - α) cos(π/2) + y cos(β - α) sin(π/2) = 0
Since cos(π/2) = 0 and sin(π/2) = 1, this simplifies to:
-x sin(β - α) + y cos(β - α) = 0
Rearranging this equation, we have:
x sin(β - α) = y cos(β - α)
Dividing both sides by cos(β - α), we get:
x tan(β - α) = y
From this equation, we can see that the slope of the perpendicular is tan(β - α).
Since the slope of the line joining the two given points is (a sin β - a sin α) / (a cos β - a cos α), we can say that:
tan(θ) = tan(β - α) = (a sin β - a sin α) / (a cos β - a cos α)
Hence, we have proven the required expression for the length of the perpendicular from the origin to the line:
Length of perpendicular = a cos((α + β)/2)