The line, L, makes an angle of 30 degrees with the positive direction of the x-axis. Find the equation of the line perpendicular to L, passing through (0,-4).

What is the question asking us? -To find the equation of a straight line, and to do that we need two pieces of information, what are they? 1. the gradient of the line and 2. a point on the line. Now from the question we are given a point on the line, so we need to find the gradient. Let's draw out the information given to us in the question. (Diagram containing information from question). There is a connection between gradient of a line and the angles they make with the positive direction of the x-axis that we can use here to find the gradient, this is m=tan(theta). (Diagram of this). When we know the angle, we can calculate the gradient and vise versa. In our case we have m=tan30. The answer to this is one that higher students are expected to know, and there is a helpful triangle we can remember and use to figure this out. (Diagram of this). So we end up with m = 1/(sqrt3). We now have the gradient of line L, but we need the gradient of the line perpendicular to L. This can be done by using the formula m_1 x m_2 = -1. So we have that the gradient of the perpendicular line is -(sqrt3). Finally we can slot this information into the general formula for a straight line which is y=mx+c to obtain y = -(sqrt3)x-4, our final answer.