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Chapter 14: Problem 1218

\(\sqrt{(3) x+y}=2\) is the equation of line containing one of the sides of an equilateral triangle and if \((0,-1)\) is one of the vertices, then the length of the side of the triangle is \(\ldots \ldots\) (a) \(\sqrt{3}\) (b) \(2 \sqrt{3}\) (c) \((\sqrt{3} / 2)\) (d) \((2 / \sqrt{3})\)

Short Answer

Expert verified
The length of the side of the equilateral triangle is \(\frac{\sqrt{3}}{2}\) which corresponds to answer choice (c).

Step by step solution

01

Find the coordinates of the other two vertices

Rewrite the given equation in the proper form of a line: \(\sqrt{3}x + y = 2\) Since (0, -1) is the first vertex, let's find the coordinates of the second vertex. The angle between the line and the x-axis is 60 degrees. We can use the slope of the line to find the other two vertices of the equilateral triangle.
02

Compute the slope and find the second vertex

The slope of the line is: \(m = - \frac{\sqrt{3}}{3}\) Now, we need to find the second vertex using this slope: \((x_2, y_2) = (0 + \frac{L}{\sqrt{3}}, -1 + L)\) where L is the length of the side of the triangle.
03

Find the third vertex

Using the properties of an equilateral triangle, we can find the third vertex: \((x_3, y_3) = (0 + 2L/\sqrt{3}, -1 - L/2)\)
04

Use the distance formula to find L

Now, we can use the distance formula to find the length of the side: \(L = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) Plugging the coordinates of the vertices, we get: \(L = \sqrt{(\frac{L}{\sqrt{3}})^2 + (-1 + L)^2}\) This simplifies to: \(\sqrt{L^2 + 2L^2 - 2L + 1} = L\)
05

Solve for L

By squaring both sides, we get: \(L^2 + 2L^2 - 2L + 1 = L^2\) Which simplifies to: \(2L^2 - 2L + 1 = 0\) Solving for L, we get: \(L = \frac{\sqrt{3}}{2}\)
06

Check the answer choices

Given the list of answers, this corresponds to answer choice (c) \((\sqrt{3} / 2)\). Therefore, the length of the side of the triangle is \((\sqrt{3} / 2)\).

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Most popular questions from this chapter

A straight line through the point \((2,2)\) intersects the line \(\sqrt{3 x}+y=0\) and \(\sqrt{3 x}-y=0\) at the points \(\mathrm{A}\) and \(B\). The equation to the line \(\mathrm{AB}\) so that the triangle \(\mathrm{OAB}\) is equilateral is (a) \(x=2\) (b) \(\mathrm{y}=2\) (c) \(\mathrm{x}+\mathrm{y}=4\) (d) none

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