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.

AS
Answered by Amy S. Maths tutor

7606 Views

See similar Maths Scottish Highers tutors

Related Maths Scottish Highers answers

All answers ▸

Given g(x) = 4* sin (3*x), find the value of g'(pi/3).


A circle has equation x^2+y^2+6x+10y-7=0. Find the equation of the tangent line through the point on the circle (-8,-1).


a) Factorise: 2x^2-72, and hence b) find the y-intercept of the line with the equation: y=(2x^2-72)/(4x-24)


Solve algebraically the following system of equations: 4x + 5y = -3; 6x - 2y = 5


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning