what is the locus of the point of intersection of the lines
xcos θ+ ysin θ= a , xsin θ - ycos θ =a
I have the answer but I dont understand the method
Hello,
$\displaystyle x \cdot \cos(\theta) + y \cdot \sin(\theta) = a$
$\displaystyle x \cdot \sin(\theta) - y \cdot \cos(\theta) = a$
Square bot sides of both equations:
$\displaystyle x^2 \cdot \cos^2(\theta) + 2 \cdot x \cdot y \cdot \sin(\theta)\cdot \cos(\theta) + y^2 \cdot \sin^2(\theta) = a^2$
$\displaystyle x^2 \cdot \sin^2(\theta) - 2 \cdot x \cdot y \cdot \sin(\theta)\cdot \cos(\theta) + y^2 \cdot \cos^2(\theta) = a^2$
Now add the columns:
$\displaystyle x^2 \cdot (\sin^2(\theta)+ \cos^2(\theta)) + y^2 \cdot (\sin^2(\theta)+ \cos^2(\theta)) = 2a^2$
$\displaystyle x^2 + y^2 = 2a^2$
This equation describes a circle with the centre at the origin and the radius $\displaystyle r = a \cdot \sqrt{2}$
Hello,
I set a = 1 and then I've drawn a few lines:
1. The dotted red lines belong to the first equations;
2. the blue lines belong to the second equation.
Have a thorough look on the points of intersection between the red and the blue lines which belong to the same angle $\displaystyle \theta$: They all lie on the black circle. That's the locus you are looking for.
(The circle which appears in the centre of this diagram is called an envelope. In this case it is the area of the coordinate plane where no line with the given properties can be.)