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Chapter 14: Problem 1257

If \(2 \mathrm{x}+2 \mathrm{y}-5=0\) is the equation of the line containing one of the sides of an equilateral triangle and \((1,2)\) is one vertex, then find the equations of the lines containing the other two sides. (a) \(\mathrm{y}=(2+\sqrt{3}) \mathrm{x}-\sqrt{3}, \mathrm{y}=(2+\sqrt{3}) \mathrm{x}+\sqrt{3}\) (b) \(y=(2-\sqrt{3}) x-\sqrt{3}, y=(2+\sqrt{3}) x+\sqrt{3}\) (c) \(y=(2-\sqrt{3}) x+\sqrt{3}, y=(2+\sqrt{3}) x-\sqrt{3}\) (d) \(y=(2+\sqrt{3}) x+\sqrt{3}, y=(2+\sqrt{3}) x-\sqrt{3}\)

Short Answer

Expert verified
The short answer is: (c) \(y = (2 - \sqrt{3})x + \sqrt{3}\) and \(y = (2 + \sqrt{3})x - \sqrt{3}\).

Step by step solution

01

Find the slope of the given equation

First, rewrite the given equation in the form y = mx + b, where m is the slope and b is the y-intercept. The given equation is: \(2x + 2y - 5 = 0\) Rewrite it to find y: \(y = -x + \frac{5}{2}\) Thus, the slope of the given equation is -1.
02

Find the slopes of the other two sides

To find the slope of the other two sides, we need to consider a 60-degree angle with the given side. The tangent of the angle between two lines with slopes m1 and m2 is given by the formula: \(\tan{\theta} = \frac{m1 - m2}{1 + m1 \cdot m2}\) For a 60-degree angle, \(\tan(60) = \sqrt{3}\). Substitute \(m1 = -1\) (slope of the given side) into the formula: \(\sqrt{3} = \frac{-1 - m2}{1 - m2}\) Solve for m2, which gives two possible slopes: \(m2 = 2 - \sqrt{3} \text{ or } m2 = 2 + \sqrt{3}\)
03

Find the equations of the other two sides

Given the slopes m2 and the given vertex (1, 2), we can find the point-slope form of the lines containing the other two sides: For m2 = \(2 - \sqrt{3}\) \(y - 2 = (2 - \sqrt{3})(x - 1)\) Which simplifies to: \(y = (2 - \sqrt{3})x + \sqrt{3}\) And for m2 = \(2 + \sqrt{3}\) \(y - 2 = (2 + \sqrt{3})(x - 1)\) Which simplifies to: \(y = (2 + \sqrt{3})x - \sqrt{3}\) So, the equations of the lines containing the other two sides are: \(y = (2 - \sqrt{3})x + \sqrt{3}\) and \(y = (2 + \sqrt{3})x - \sqrt{3}\) And the answer is (c).

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Most popular questions from this chapter

The length of a side of a square OPQR is \(\mathrm{a}, \mathrm{O}\) is the origin OP and \(\mathrm{OR}\) are along positive direction of the \(\mathrm{X}\) and \(\mathrm{Y}\) axes respectively. If \(\mathrm{A}\) and \(\mathrm{B}\) are mid points of \(\underline{\mathrm{PQ}}\) and \(\underline{\mathrm{QR}}\) respectively then measure of angle between \(\underline{\mathrm{OA}}\) and \(\underline{\mathrm{OB}}\) is.... (a) \(\cos ^{-1}(3 / 5)\) (b) \(\tan ^{-1}(4 / 3)\) (c) \(\cot ^{-1}(3 / 4)\) (d) \(\sin ^{-1}(3 / 5)\)

If a vertex of a triangle is \((1,1)\) and the mid-points of two sides through this vertex are \((-1,2)\) and \((3,2)\), then centroid of the triangle is (a) \([(1 / 3),(7 / 3)]\) (b) \([1,(7 / 3)]\) (c) \([-(1 / 3),(7 / 3)]\) (d) \([-1,(7 / 3)]\)

The equation of line passing through the point \((1,2)\) and making the intercept of length 3 units between the lines \(3 \mathrm{x}+4 \mathrm{y}=24\) and \(3 \mathrm{x}+4 \mathrm{y}=12\), is \(\ldots \ldots\) (a) \(7 \mathrm{x}-24 \mathrm{y}+41=0\) (b) \(7 x+24 y=55\) (c) \(24 x-7 y=10\) (d) \(24 x+7 y-38=0\)

A line intersects \(\mathrm{X}\) -axis and Y-axis at \(\mathrm{A}\) and \(\mathrm{B}\) respectively. If \(\mathrm{AB}=15\) and \(\underline{\mathrm{AB}}\) makes a triangle of area 54 units with coordinate axes, then the equation of \(\underline{A B}\) is \(\ldots .\) (a) \(4 x \pm 3 y=36\) or \(3 x \pm 4 y=36\) (b) \(4 x \pm 3 y=24\) or \(3 x \pm 4 y=24\) (c) \(-4 \mathrm{x} \pm 3 \mathrm{y}=24\) or \(-3 \mathrm{x} \pm 4 \mathrm{y}=24\) (d) \(-4 x \pm 3 y=12\) or \(-3 x \pm 4 y=12\)

The length of side of an equilateral triangle is a. There is circle inscribed in a triangle. What is the area of a square inscribed in a circle ? (a) \(\left(\mathrm{a}^{2} / 3\right)\) (b) \(\left(\mathrm{a}^{2} / 6\right)\) (c) \(\left(\mathrm{a}^{2} / \sqrt{3}\right)\) (d) \(\left(a^{2} / \sqrt{2}\right)\)
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