To find the distance between two points $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$, we use the distance formula:
$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $
In this case, the points are:
Let's substitute the coordinates into the distance formula:
$ d = \sqrt{(0 - a \cos \alpha)^2 + (a \sin \alpha - 0)^2} $
Simplify the terms inside the square root:
$ d = \sqrt{(-a \cos \alpha)^2 + (a \sin \alpha)^2} $
$ d = \sqrt{a^2 \cos^2 \alpha + a^2 \sin^2 \alpha} $
Factor out $a^2$:
$ d = \sqrt{a^2 (\cos^2 \alpha + \sin^2 \alpha)} $
Using the trigonometric identity $\cos^2 \alpha + \sin^2 \alpha = 1$:
$ d = \sqrt{a^2 \times 1} $
$ d = \sqrt{a^2} $
The square root of $a^2$ is the absolute value of $a$.
$ d = |a| $
Therefore, the distance between the two given points is $|a|$.
Find the co-ordinates of the centroid of a triangle whose vertices are A(1, 4), B(7, 8) and C(10, 12).