Question 665604: Two lines intersect at (9,-4). This point, together with the y-intercepts of the two lines, are the vertices of an equilateral triangle. Find the coordinates of the other two vertices.
Answer by DrBeeee(684) (Show Source):
You can put this solution on YOUR website! This is tricky so follow carefully, because I can't draw pretty pictures!
First visualize the given point (9,-4), 9 units to the right of the origin and 4 units below the x-axis. Since this is a vertex of an equilateral triangle formed by this point and the two y-intercepts (where x = 0), the height of the triangle is 9 units. Correct?
We also know that all of the angles of an equilateral are 60 degrees each and that the perpendicular bisector of the vertex is at 30 degrees.
So visualize a right triangle whose base is on the y-axis, with its other leg parallel to the x-axis that intersects the y-axis at -4 units and ends at the point (9,-4). The hypotenuse of this triangle goes from the vertex point (9,-4) back to the y-axis at an angle of 30 degrees to the horizontal leg. The length of this hypotenuse is given by
(1) h = 9/cos(30) or
(2) h = 18/sqrt(3)
This value is the lenght of the sides of the equilateral triangle. The base leg of the right triangle is one half of this or 9/sqrt(3). The third side runs along the y-axis for a length of 18/sqrt(3), half of which is above y = -4 and the other half is below y = -4. Since the x coordinate is zero on the y-axis, the other vertices are at
(3) (0,-4-9/sqrt(3)) and
(4) (0,-4+9/sqrt(3))
Answer: The coordinates of the other two vertices are given by (3) and (4).
|
|
|