1.The line passing through (−1,π2)(-1, \frac{\pi}{2})(−1,2π) and perpendicular to 3sinθ+2cosθ=4r\sqrt{3}\sin\theta + 2\cos\theta = \frac{4}{r}3sinθ+2cosθ=r4 isa.2=3rcosθ−2rsinθ2 = \sqrt{3}r\cos\theta - 2r\sin\theta2=3rcosθ−2rsinθb.5=−23rsinθ+4rcosθ5 = -2\sqrt{3}r\sin\theta + 4r\cos\theta5=−23rsinθ+4rcosθc.2=3rcosθ+2rsinθ2 = \sqrt{3}r\cos\theta + 2r\sin\theta2=3rcosθ+2rsinθd.5=23rsinθ+4rcosθ5 = 2\sqrt{3}r\sin\theta + 4r\cos\theta5=23rsinθ+4rcosθLogin to continueOnly logged in users canattempt or see the solution.
