Find the joint equation of pair of lines through
the origin and making an equilateral triangle
with the line y = 3
Answers
Answer
x=3 is a vertical line (shown in blue in the figure).
We need to add two lines so the three lines will form an equilateral triangle. Since the two new lines are both supposed to pass through the origin, it must be one of the vertices of the triangle, and the other two vertices will be intersections of the new lines with x=3.
But since the line from the origin to x=3 (that is, the x-axis) is perpendicular to it, the x-axis must be an altitude of the equilateral triangle. That means two things:
(1) An altitude of an equilateral triangle is also a median, so point (3,0), where x=3 crosses the x-axis, is a midpoint between the other two vertices.
(2) We know the altitude of our equilateral triangle measures exactly 3, and since the altitude of an equilateral triangle measures (√3)/2 times the length of a side, we know the sides of the triangle must each have a length of
3 / [(√3)/2] = 2√3
From these, we see the other two vertices of the triangle must be on x=3, one √3 above the x-axis and one √3 below it. So they are the points
(3,√3) and (3, -√3)
The y-intercept of each line is, of course, the origin.
And the slope of the two lines is
(difference in y-coordinates) / (difference in x-coordinates)
when comparing the other known points on the line with the origin (but the differences are just the coordinate values of the points). So the slopes of the lines are
(√3)/3 and -(√3)/3
Thus, their equations are easy to write in slope-intercept form:
y = mx + b
where m is the slope and b is the y-intercept (which we need not write because it's zero).
The pair of lines have the equations
y = (√3)x/3 and
y = -(√3)x/3
But we want a JOINT equation for the pair of lines. We can obtain that by eliminating the distinction between positive and negative values of y, either writing it as
y = ±(√3)x/3
or [and this is what I suspect your instructor expects] as
|y| = (√3)x/3
where |y| is the absolute value of y.