Answer to Question #145008 in Analytic Geometry for Sarita bartwal

Question

Does the equation 2/r= 3cos(θ-π/4)+ 2sin(θ+π/4) represent a straight line? Satisfy your answer

Answer 1

$\displaystyle \frac{2}{r}= 3\cos\left(\theta -\frac{\pi}{4}\right) + 2\sin\left(\theta +\frac{\pi}{4}\right)\\ r = \frac{2}{3\cos\left(\theta -\frac{\pi}{4}\right) + 2\sin\left(\theta +\frac{\pi}{4}\right)}\\ \cos\left(\frac{\pi}{2} - \theta\right) = \sin{\theta}\\ r = \frac{2}{3\sin\left(-\theta + \frac{3\pi}{4}\right) + 2\sin\left(\theta +\frac{\pi}{4}\right)}\\ \sin{\theta} = \sin(\pi - \theta)\\ r = \frac{2}{3\sin\left(\theta + \frac{\pi}{4}\right) + 2\sin\left(\theta +\frac{\pi}{4}\right)}\\ r = \frac{2}{5\sin\left(\theta + \frac{\pi}{4}\right)}\\ r^2 = \frac{4}{25\sin^2\left(\theta + \frac{\pi}{4}\right)}\\ x^2 + y^2 = \frac{4}{25\sin^2\left(\theta + \frac{\pi}{4}\right)}\\ x^2 + y^2 = \frac{8}{25\left(1 - \cos\left(2\theta + \frac{\pi}{2}\right)\right)}\\ x^2 + y^2 = \frac{8}{25\left(1 + \sin(2\theta)\right)}\\ x^2 + y^2 = \frac{8}{25\left(1 + 2\sin{\theta}\cos{\theta}\right)}\\ x = r\cos\theta, y = r\sin\theta\\ x^2 + y^2 = \frac{8}{25\left(1 + \frac{2xy}{r^2} \right)}\\ x^2 + y^2 = \frac{8(x^2 + y^2)}{25\left(x^2 + y^2 + 2xy \right)}\\ 1 = \frac{8}{25\left(x^2 + y^2 + 2xy \right)}\\ x^2 + y^2 + 2xy = \frac{8}{25}\\ (x + y)^2 = \frac{8}{25}\\ x + y = \pm\frac{2\sqrt{2}}{5}\\ \textsf{This is of course the}\\ \textsf{equation of a straight line.}\\ \textsf{A plot is attached above}$